143766
A body projected vertically upwards crosses a point twice in its journey at a height \(h\) just after \(t_{1}\) and \(t_{2}\) second. Maximum height reached by the body is
C We know, \(\mathrm{h}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\) At \(\mathrm{t}=\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) second Then, \(\mathrm{h}=\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}\) \(\mathrm{h}=\mathrm{ut}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2}\) Equating (i) and (ii), we get- \(\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}=\mathrm{ut}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2}\) \(\mathrm{u}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}^{2}-\mathrm{t}_{2}^{2}\right)\) \(\mathrm{u}=\frac{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right) \mathrm{g}}{2}\) Maximum height \(=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{\mathrm{g}^{2}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}}{4 \times 2 \mathrm{~g}}=2 \mathrm{~g}\left(\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{4}\right)^{2}\)
AP EAMCET-2005
Motion in Plane
143769
A ball thrown vertically up to reach its maximum height in \(t\) second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is
B At maximum height final velocity is zero. So, Initial velocity \(\left(\mathrm{v}_{0}\right)=\sqrt{2 \mathrm{gH}}\) And, \(\quad \mathrm{t}=\frac{\mathrm{v}_{0}}{\mathrm{~g}}\) \(\mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} .\) Applying the relation for the motion from the point A and \(\mathrm{C}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}_{\mathrm{C}}^{2}=\mathrm{v}_{\mathrm{A}}^{2}+2 \mathrm{as}\) \(\mathrm{v}_{\mathrm{C}}^{2}=0+2 \mathrm{~g}\left(\frac{\mathrm{H}}{2}\right)\) \(\mathrm{v}_{\mathrm{C}}=\sqrt{\mathrm{gH}} \ldots \ldots .\) For motion \(\mathrm{A}\) to \(\mathrm{C}\) \(\mathrm{v}_{\mathrm{C}}=\mathrm{u}_{\mathrm{A}}+\mathrm{at}_{1}\) \(\mathrm{v}_{\mathrm{C}}=\mathrm{gt}_{1} \ldots \ldots \ldots \ldots . .(\mathrm{v})\) \(\sqrt{\mathrm{gH}}=\mathrm{gt}_{1} \quad \text { [From (iv)] }\) \(\mathrm{t}_{1}=\sqrt{\frac{\mathrm{H}}{\mathrm{g}}} \ldots \ldots \ldots \ldots(\mathrm{vi})\) Comparing equation (iii) and (vi) \(\mathrm{t}_{1}=\frac{\mathrm{t}}{\sqrt{2}}\) So, the time taken to reach half of the maximum height while returning is equal to \(\mathrm{t}_{2}=\mathrm{t}+\mathrm{t}_{1}\) \(\mathrm{t}_{2}=\mathrm{t}+\frac{\mathrm{t}}{\sqrt{2}}\) \(\mathrm{t}_{2}=\left(1+\frac{1}{\sqrt{2}}\right) \mathrm{t}\)
AP EAMCET-2008
Motion in Plane
143770
For an object thrown at \(45^{\circ}\) to horizontal, the maximum height \((\mathrm{H})\) and horizontal range \((\mathrm{R})\) are related as
1 \(\mathrm{R}=16 \mathrm{H}\)
2 \(\mathrm{R}=8 \mathrm{H}\)
3 \(\mathrm{R}=4 \mathrm{H}\)
4 \(\mathrm{R}=2 \mathrm{H}\)
Explanation:
C Given, \(\theta=45^{\circ}\) Maximum height of projectile \(\left(\mathrm{H}_{\max }\right)=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) \(\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} 45^{\circ}}{2 \mathrm{~g}}\) \(\text { Range of projectile }(\mathrm{R})=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin \left(2 \times 45^{\circ}\right)}{\mathrm{g}}\) \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin 90^{\circ}}{\mathrm{g}} \ldots \ldots . \text { (ii) }\) Dividing Equation (i) by Equation (ii), we get- \(\frac{\mathrm{H}_{\max }}{\mathrm{R}}=\frac{\mathrm{u}^{2} \sin ^{2} 45^{\circ} \times \mathrm{g}}{2 \mathrm{~g} \times \mathrm{u}^{2} \sin 90^{\circ}}=\frac{1}{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{4}\) So, \(\quad \mathrm{R}=4 \mathrm{H}_{\max }\)
DCE-2009
Motion in Plane
143771
Two objects are projected at an angle \(\theta\) and \(\left(90^{\circ}-\theta\right)\), to the horizontal with the same speed. The ratio of their maximum vertical heights is
1 \(1: \tan \theta\)
2 \(\tan ^{2} \theta: 1\)
3 \(1: 1\)
4 \(\tan \theta: 1\)
Explanation:
B When object projected at an angle \(\theta\) then, \(\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Now when object projected at an angle \(\left(90^{\circ}-\theta\right)\) then, \(\mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}\) Now divide equation (i) by (ii), we get- \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}}{\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}}\) \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{\mathrm{u}^{2} \cos ^{2} \theta}\) \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\tan ^{2} \theta}{1}\) \(\mathrm{H}_{1}: \mathrm{H}_{2}=\tan ^{2} \theta: 1\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in Plane
143766
A body projected vertically upwards crosses a point twice in its journey at a height \(h\) just after \(t_{1}\) and \(t_{2}\) second. Maximum height reached by the body is
C We know, \(\mathrm{h}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\) At \(\mathrm{t}=\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) second Then, \(\mathrm{h}=\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}\) \(\mathrm{h}=\mathrm{ut}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2}\) Equating (i) and (ii), we get- \(\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}=\mathrm{ut}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2}\) \(\mathrm{u}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}^{2}-\mathrm{t}_{2}^{2}\right)\) \(\mathrm{u}=\frac{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right) \mathrm{g}}{2}\) Maximum height \(=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{\mathrm{g}^{2}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}}{4 \times 2 \mathrm{~g}}=2 \mathrm{~g}\left(\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{4}\right)^{2}\)
AP EAMCET-2005
Motion in Plane
143769
A ball thrown vertically up to reach its maximum height in \(t\) second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is
B At maximum height final velocity is zero. So, Initial velocity \(\left(\mathrm{v}_{0}\right)=\sqrt{2 \mathrm{gH}}\) And, \(\quad \mathrm{t}=\frac{\mathrm{v}_{0}}{\mathrm{~g}}\) \(\mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} .\) Applying the relation for the motion from the point A and \(\mathrm{C}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}_{\mathrm{C}}^{2}=\mathrm{v}_{\mathrm{A}}^{2}+2 \mathrm{as}\) \(\mathrm{v}_{\mathrm{C}}^{2}=0+2 \mathrm{~g}\left(\frac{\mathrm{H}}{2}\right)\) \(\mathrm{v}_{\mathrm{C}}=\sqrt{\mathrm{gH}} \ldots \ldots .\) For motion \(\mathrm{A}\) to \(\mathrm{C}\) \(\mathrm{v}_{\mathrm{C}}=\mathrm{u}_{\mathrm{A}}+\mathrm{at}_{1}\) \(\mathrm{v}_{\mathrm{C}}=\mathrm{gt}_{1} \ldots \ldots \ldots \ldots . .(\mathrm{v})\) \(\sqrt{\mathrm{gH}}=\mathrm{gt}_{1} \quad \text { [From (iv)] }\) \(\mathrm{t}_{1}=\sqrt{\frac{\mathrm{H}}{\mathrm{g}}} \ldots \ldots \ldots \ldots(\mathrm{vi})\) Comparing equation (iii) and (vi) \(\mathrm{t}_{1}=\frac{\mathrm{t}}{\sqrt{2}}\) So, the time taken to reach half of the maximum height while returning is equal to \(\mathrm{t}_{2}=\mathrm{t}+\mathrm{t}_{1}\) \(\mathrm{t}_{2}=\mathrm{t}+\frac{\mathrm{t}}{\sqrt{2}}\) \(\mathrm{t}_{2}=\left(1+\frac{1}{\sqrt{2}}\right) \mathrm{t}\)
AP EAMCET-2008
Motion in Plane
143770
For an object thrown at \(45^{\circ}\) to horizontal, the maximum height \((\mathrm{H})\) and horizontal range \((\mathrm{R})\) are related as
1 \(\mathrm{R}=16 \mathrm{H}\)
2 \(\mathrm{R}=8 \mathrm{H}\)
3 \(\mathrm{R}=4 \mathrm{H}\)
4 \(\mathrm{R}=2 \mathrm{H}\)
Explanation:
C Given, \(\theta=45^{\circ}\) Maximum height of projectile \(\left(\mathrm{H}_{\max }\right)=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) \(\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} 45^{\circ}}{2 \mathrm{~g}}\) \(\text { Range of projectile }(\mathrm{R})=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin \left(2 \times 45^{\circ}\right)}{\mathrm{g}}\) \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin 90^{\circ}}{\mathrm{g}} \ldots \ldots . \text { (ii) }\) Dividing Equation (i) by Equation (ii), we get- \(\frac{\mathrm{H}_{\max }}{\mathrm{R}}=\frac{\mathrm{u}^{2} \sin ^{2} 45^{\circ} \times \mathrm{g}}{2 \mathrm{~g} \times \mathrm{u}^{2} \sin 90^{\circ}}=\frac{1}{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{4}\) So, \(\quad \mathrm{R}=4 \mathrm{H}_{\max }\)
DCE-2009
Motion in Plane
143771
Two objects are projected at an angle \(\theta\) and \(\left(90^{\circ}-\theta\right)\), to the horizontal with the same speed. The ratio of their maximum vertical heights is
1 \(1: \tan \theta\)
2 \(\tan ^{2} \theta: 1\)
3 \(1: 1\)
4 \(\tan \theta: 1\)
Explanation:
B When object projected at an angle \(\theta\) then, \(\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Now when object projected at an angle \(\left(90^{\circ}-\theta\right)\) then, \(\mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}\) Now divide equation (i) by (ii), we get- \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}}{\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}}\) \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{\mathrm{u}^{2} \cos ^{2} \theta}\) \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\tan ^{2} \theta}{1}\) \(\mathrm{H}_{1}: \mathrm{H}_{2}=\tan ^{2} \theta: 1\)
143766
A body projected vertically upwards crosses a point twice in its journey at a height \(h\) just after \(t_{1}\) and \(t_{2}\) second. Maximum height reached by the body is
C We know, \(\mathrm{h}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\) At \(\mathrm{t}=\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) second Then, \(\mathrm{h}=\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}\) \(\mathrm{h}=\mathrm{ut}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2}\) Equating (i) and (ii), we get- \(\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}=\mathrm{ut}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2}\) \(\mathrm{u}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}^{2}-\mathrm{t}_{2}^{2}\right)\) \(\mathrm{u}=\frac{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right) \mathrm{g}}{2}\) Maximum height \(=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{\mathrm{g}^{2}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}}{4 \times 2 \mathrm{~g}}=2 \mathrm{~g}\left(\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{4}\right)^{2}\)
AP EAMCET-2005
Motion in Plane
143769
A ball thrown vertically up to reach its maximum height in \(t\) second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is
B At maximum height final velocity is zero. So, Initial velocity \(\left(\mathrm{v}_{0}\right)=\sqrt{2 \mathrm{gH}}\) And, \(\quad \mathrm{t}=\frac{\mathrm{v}_{0}}{\mathrm{~g}}\) \(\mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} .\) Applying the relation for the motion from the point A and \(\mathrm{C}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}_{\mathrm{C}}^{2}=\mathrm{v}_{\mathrm{A}}^{2}+2 \mathrm{as}\) \(\mathrm{v}_{\mathrm{C}}^{2}=0+2 \mathrm{~g}\left(\frac{\mathrm{H}}{2}\right)\) \(\mathrm{v}_{\mathrm{C}}=\sqrt{\mathrm{gH}} \ldots \ldots .\) For motion \(\mathrm{A}\) to \(\mathrm{C}\) \(\mathrm{v}_{\mathrm{C}}=\mathrm{u}_{\mathrm{A}}+\mathrm{at}_{1}\) \(\mathrm{v}_{\mathrm{C}}=\mathrm{gt}_{1} \ldots \ldots \ldots \ldots . .(\mathrm{v})\) \(\sqrt{\mathrm{gH}}=\mathrm{gt}_{1} \quad \text { [From (iv)] }\) \(\mathrm{t}_{1}=\sqrt{\frac{\mathrm{H}}{\mathrm{g}}} \ldots \ldots \ldots \ldots(\mathrm{vi})\) Comparing equation (iii) and (vi) \(\mathrm{t}_{1}=\frac{\mathrm{t}}{\sqrt{2}}\) So, the time taken to reach half of the maximum height while returning is equal to \(\mathrm{t}_{2}=\mathrm{t}+\mathrm{t}_{1}\) \(\mathrm{t}_{2}=\mathrm{t}+\frac{\mathrm{t}}{\sqrt{2}}\) \(\mathrm{t}_{2}=\left(1+\frac{1}{\sqrt{2}}\right) \mathrm{t}\)
AP EAMCET-2008
Motion in Plane
143770
For an object thrown at \(45^{\circ}\) to horizontal, the maximum height \((\mathrm{H})\) and horizontal range \((\mathrm{R})\) are related as
1 \(\mathrm{R}=16 \mathrm{H}\)
2 \(\mathrm{R}=8 \mathrm{H}\)
3 \(\mathrm{R}=4 \mathrm{H}\)
4 \(\mathrm{R}=2 \mathrm{H}\)
Explanation:
C Given, \(\theta=45^{\circ}\) Maximum height of projectile \(\left(\mathrm{H}_{\max }\right)=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) \(\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} 45^{\circ}}{2 \mathrm{~g}}\) \(\text { Range of projectile }(\mathrm{R})=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin \left(2 \times 45^{\circ}\right)}{\mathrm{g}}\) \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin 90^{\circ}}{\mathrm{g}} \ldots \ldots . \text { (ii) }\) Dividing Equation (i) by Equation (ii), we get- \(\frac{\mathrm{H}_{\max }}{\mathrm{R}}=\frac{\mathrm{u}^{2} \sin ^{2} 45^{\circ} \times \mathrm{g}}{2 \mathrm{~g} \times \mathrm{u}^{2} \sin 90^{\circ}}=\frac{1}{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{4}\) So, \(\quad \mathrm{R}=4 \mathrm{H}_{\max }\)
DCE-2009
Motion in Plane
143771
Two objects are projected at an angle \(\theta\) and \(\left(90^{\circ}-\theta\right)\), to the horizontal with the same speed. The ratio of their maximum vertical heights is
1 \(1: \tan \theta\)
2 \(\tan ^{2} \theta: 1\)
3 \(1: 1\)
4 \(\tan \theta: 1\)
Explanation:
B When object projected at an angle \(\theta\) then, \(\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Now when object projected at an angle \(\left(90^{\circ}-\theta\right)\) then, \(\mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}\) Now divide equation (i) by (ii), we get- \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}}{\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}}\) \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{\mathrm{u}^{2} \cos ^{2} \theta}\) \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\tan ^{2} \theta}{1}\) \(\mathrm{H}_{1}: \mathrm{H}_{2}=\tan ^{2} \theta: 1\)
143766
A body projected vertically upwards crosses a point twice in its journey at a height \(h\) just after \(t_{1}\) and \(t_{2}\) second. Maximum height reached by the body is
C We know, \(\mathrm{h}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\) At \(\mathrm{t}=\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) second Then, \(\mathrm{h}=\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}\) \(\mathrm{h}=\mathrm{ut}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2}\) Equating (i) and (ii), we get- \(\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}=\mathrm{ut}_{2}-\frac{1}{2} \mathrm{gt}_{2}^{2}\) \(\mathrm{u}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}^{2}-\mathrm{t}_{2}^{2}\right)\) \(\mathrm{u}=\frac{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right) \mathrm{g}}{2}\) Maximum height \(=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{\mathrm{g}^{2}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}}{4 \times 2 \mathrm{~g}}=2 \mathrm{~g}\left(\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{4}\right)^{2}\)
AP EAMCET-2005
Motion in Plane
143769
A ball thrown vertically up to reach its maximum height in \(t\) second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is
B At maximum height final velocity is zero. So, Initial velocity \(\left(\mathrm{v}_{0}\right)=\sqrt{2 \mathrm{gH}}\) And, \(\quad \mathrm{t}=\frac{\mathrm{v}_{0}}{\mathrm{~g}}\) \(\mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} .\) Applying the relation for the motion from the point A and \(\mathrm{C}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(\mathrm{v}_{\mathrm{C}}^{2}=\mathrm{v}_{\mathrm{A}}^{2}+2 \mathrm{as}\) \(\mathrm{v}_{\mathrm{C}}^{2}=0+2 \mathrm{~g}\left(\frac{\mathrm{H}}{2}\right)\) \(\mathrm{v}_{\mathrm{C}}=\sqrt{\mathrm{gH}} \ldots \ldots .\) For motion \(\mathrm{A}\) to \(\mathrm{C}\) \(\mathrm{v}_{\mathrm{C}}=\mathrm{u}_{\mathrm{A}}+\mathrm{at}_{1}\) \(\mathrm{v}_{\mathrm{C}}=\mathrm{gt}_{1} \ldots \ldots \ldots \ldots . .(\mathrm{v})\) \(\sqrt{\mathrm{gH}}=\mathrm{gt}_{1} \quad \text { [From (iv)] }\) \(\mathrm{t}_{1}=\sqrt{\frac{\mathrm{H}}{\mathrm{g}}} \ldots \ldots \ldots \ldots(\mathrm{vi})\) Comparing equation (iii) and (vi) \(\mathrm{t}_{1}=\frac{\mathrm{t}}{\sqrt{2}}\) So, the time taken to reach half of the maximum height while returning is equal to \(\mathrm{t}_{2}=\mathrm{t}+\mathrm{t}_{1}\) \(\mathrm{t}_{2}=\mathrm{t}+\frac{\mathrm{t}}{\sqrt{2}}\) \(\mathrm{t}_{2}=\left(1+\frac{1}{\sqrt{2}}\right) \mathrm{t}\)
AP EAMCET-2008
Motion in Plane
143770
For an object thrown at \(45^{\circ}\) to horizontal, the maximum height \((\mathrm{H})\) and horizontal range \((\mathrm{R})\) are related as
1 \(\mathrm{R}=16 \mathrm{H}\)
2 \(\mathrm{R}=8 \mathrm{H}\)
3 \(\mathrm{R}=4 \mathrm{H}\)
4 \(\mathrm{R}=2 \mathrm{H}\)
Explanation:
C Given, \(\theta=45^{\circ}\) Maximum height of projectile \(\left(\mathrm{H}_{\max }\right)=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) \(\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} 45^{\circ}}{2 \mathrm{~g}}\) \(\text { Range of projectile }(\mathrm{R})=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin \left(2 \times 45^{\circ}\right)}{\mathrm{g}}\) \(\mathrm{R}=\frac{\mathrm{u}^{2} \sin 90^{\circ}}{\mathrm{g}} \ldots \ldots . \text { (ii) }\) Dividing Equation (i) by Equation (ii), we get- \(\frac{\mathrm{H}_{\max }}{\mathrm{R}}=\frac{\mathrm{u}^{2} \sin ^{2} 45^{\circ} \times \mathrm{g}}{2 \mathrm{~g} \times \mathrm{u}^{2} \sin 90^{\circ}}=\frac{1}{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{4}\) So, \(\quad \mathrm{R}=4 \mathrm{H}_{\max }\)
DCE-2009
Motion in Plane
143771
Two objects are projected at an angle \(\theta\) and \(\left(90^{\circ}-\theta\right)\), to the horizontal with the same speed. The ratio of their maximum vertical heights is
1 \(1: \tan \theta\)
2 \(\tan ^{2} \theta: 1\)
3 \(1: 1\)
4 \(\tan \theta: 1\)
Explanation:
B When object projected at an angle \(\theta\) then, \(\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Now when object projected at an angle \(\left(90^{\circ}-\theta\right)\) then, \(\mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}\) Now divide equation (i) by (ii), we get- \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}}{\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}}\) \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{\mathrm{u}^{2} \cos ^{2} \theta}\) \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\tan ^{2} \theta}{1}\) \(\mathrm{H}_{1}: \mathrm{H}_{2}=\tan ^{2} \theta: 1\)
