143805
From the top of a tower \(19.6 \mathrm{~m}\) high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of \(45^{\circ}\) with the horizontal, then the time initial velocity of the ball is
1 \(9.8 \mathrm{~ms}^{-1}\)
2 \(4.9 \mathrm{~ms}^{-1}\)
3 \(14.7 \mathrm{~ms}^{-1}\)
4 \(2.8 \mathrm{~ms}^{-1}\)
Explanation:
A Given, \(\theta=45^{\circ}, \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}\) Top of tower height \((\mathrm{H})=19.6 \mathrm{~m}\) \(\because \quad \theta=45^{\circ}\) \(R=19.6\) and \(\quad \tan \theta=\frac{H}{R}\) \(\tan 45^{\circ}=\frac{H}{R}\) \(R=H=19.6 \mathrm{~m}\) In Horizontal direction \(\mathrm{R}=19.6=\mathrm{u} \times \mathrm{t}\) In vertical direction \(\mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\) \(\mathrm{t}=\sqrt{\frac{2 \times 19.6}{9.8}}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \sec\) By putting value of \(t\) in equation (1), we get- \(19.6=\mathrm{u} \times \mathrm{t}\) \(19.6=\mathrm{u} \times 2\) \(\mathrm{u}=\frac{19.6}{2}\) \(\mathrm{u}=9.8 \mathrm{~m} / \mathrm{s}\)
MHT-CET 2020
Motion in Plane
143806
A ball is thrown at two different angles with the same speed \(v\) and from the same point and it has the same range in both the cases. If \(y_{1}\) and \(y_{2}\) be the heights attained in the two cases, then \(y_{1}+y_{2}=\ldots\) :
1 \(\mathrm{u}^{2} / \mathrm{g}\)
2 \(2 u^{2} / g\)
3 \(\mathrm{u}^{2} / 2 \mathrm{~g}\)
4 \(\mathrm{u}^{2} / 4 \mathrm{~g}\)
Explanation:
C Same range can be obtained from complementary angles i.e. \(\theta\) and \(\left(90^{\circ}-\theta\right)\) Height attained by \(\mathrm{I}^{\text {st }}\) ball \(\mathrm{y}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Height attained by II \(^{\text {nd }}\) ball \(\mathrm{y}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}(90-\theta)}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}}\) \(\therefore \quad y_{1}+y_{2}=\frac{u^{2} \sin ^{2} \theta}{2 g}+\frac{u^{2} \cos ^{2} \theta}{2 g}\) \(=\frac{u^{2} \sin ^{2} \theta+u^{2} \cos ^{2} \theta}{2 g}\) \(=\frac{u^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)}{2 g} \quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]\) \(=\frac{u^{2}}{2 g}\)
MHT-CET 2020
Motion in Plane
143807
At what angle with the horizontal should a ball be thrown so that its range \(R\) is related to the time of flight as \(R=\mathbf{5} T^{2}\). (Take \(g=10 \mathrm{~ms}^{-2}\) )
143808
The maximum height attained by a projectile is increased by \(10 \%\). Keeping the angle of projection constant, what is percentage increase in the time of flight?
1 \(5 \%\)
2 \(10 \%\)
3 \(20 \%\)
4 \(40 \%\)
Explanation:
A We know, \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Time of flight \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\). Squaring on both side of equation (2), we get- \(\mathrm{T}^{2}=\left(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\right)^{2}=\frac{4 \mathrm{u}^{2} \sin ^{2} \theta}{\mathrm{g}^{2}}\) \(\mathrm{u}^{2} \sin ^{2} \theta=\frac{\mathrm{T}^{2} \times \mathrm{g}^{2}}{4}\) Putting the value of \(u^{2} \sin ^{2} \theta\) from equation (1), we get- \(\mathrm{H} \times 2 \mathrm{~g}=\frac{\mathrm{T}^{2} \times \mathrm{g}^{2}}{4}\) \(\mathrm{H}=\mathrm{T}^{2}\left(\frac{\mathrm{g}}{8}\right)\) \(\frac{\mathrm{g}}{8} \rightarrow \text { constant }\) So, \(\frac{\Delta H}{H}=\frac{2 \Delta T}{T}\) \(\frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta H}{H}\) \(\frac{\Delta T}{T} \times 100=\frac{1}{2}\left(\frac{\Delta H}{H} \times 100\right) \quad\left(\because \frac{\Delta H}{H} \times 100=10\right)\) \(\frac{\Delta T}{T} \times 100=\frac{1}{2} \times 10\) \(\frac{\Delta T}{T} \times 100=5\) \(\%\) increases in \(\mathrm{T}=5 \%\)
143805
From the top of a tower \(19.6 \mathrm{~m}\) high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of \(45^{\circ}\) with the horizontal, then the time initial velocity of the ball is
1 \(9.8 \mathrm{~ms}^{-1}\)
2 \(4.9 \mathrm{~ms}^{-1}\)
3 \(14.7 \mathrm{~ms}^{-1}\)
4 \(2.8 \mathrm{~ms}^{-1}\)
Explanation:
A Given, \(\theta=45^{\circ}, \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}\) Top of tower height \((\mathrm{H})=19.6 \mathrm{~m}\) \(\because \quad \theta=45^{\circ}\) \(R=19.6\) and \(\quad \tan \theta=\frac{H}{R}\) \(\tan 45^{\circ}=\frac{H}{R}\) \(R=H=19.6 \mathrm{~m}\) In Horizontal direction \(\mathrm{R}=19.6=\mathrm{u} \times \mathrm{t}\) In vertical direction \(\mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\) \(\mathrm{t}=\sqrt{\frac{2 \times 19.6}{9.8}}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \sec\) By putting value of \(t\) in equation (1), we get- \(19.6=\mathrm{u} \times \mathrm{t}\) \(19.6=\mathrm{u} \times 2\) \(\mathrm{u}=\frac{19.6}{2}\) \(\mathrm{u}=9.8 \mathrm{~m} / \mathrm{s}\)
MHT-CET 2020
Motion in Plane
143806
A ball is thrown at two different angles with the same speed \(v\) and from the same point and it has the same range in both the cases. If \(y_{1}\) and \(y_{2}\) be the heights attained in the two cases, then \(y_{1}+y_{2}=\ldots\) :
1 \(\mathrm{u}^{2} / \mathrm{g}\)
2 \(2 u^{2} / g\)
3 \(\mathrm{u}^{2} / 2 \mathrm{~g}\)
4 \(\mathrm{u}^{2} / 4 \mathrm{~g}\)
Explanation:
C Same range can be obtained from complementary angles i.e. \(\theta\) and \(\left(90^{\circ}-\theta\right)\) Height attained by \(\mathrm{I}^{\text {st }}\) ball \(\mathrm{y}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Height attained by II \(^{\text {nd }}\) ball \(\mathrm{y}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}(90-\theta)}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}}\) \(\therefore \quad y_{1}+y_{2}=\frac{u^{2} \sin ^{2} \theta}{2 g}+\frac{u^{2} \cos ^{2} \theta}{2 g}\) \(=\frac{u^{2} \sin ^{2} \theta+u^{2} \cos ^{2} \theta}{2 g}\) \(=\frac{u^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)}{2 g} \quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]\) \(=\frac{u^{2}}{2 g}\)
MHT-CET 2020
Motion in Plane
143807
At what angle with the horizontal should a ball be thrown so that its range \(R\) is related to the time of flight as \(R=\mathbf{5} T^{2}\). (Take \(g=10 \mathrm{~ms}^{-2}\) )
143808
The maximum height attained by a projectile is increased by \(10 \%\). Keeping the angle of projection constant, what is percentage increase in the time of flight?
1 \(5 \%\)
2 \(10 \%\)
3 \(20 \%\)
4 \(40 \%\)
Explanation:
A We know, \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Time of flight \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\). Squaring on both side of equation (2), we get- \(\mathrm{T}^{2}=\left(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\right)^{2}=\frac{4 \mathrm{u}^{2} \sin ^{2} \theta}{\mathrm{g}^{2}}\) \(\mathrm{u}^{2} \sin ^{2} \theta=\frac{\mathrm{T}^{2} \times \mathrm{g}^{2}}{4}\) Putting the value of \(u^{2} \sin ^{2} \theta\) from equation (1), we get- \(\mathrm{H} \times 2 \mathrm{~g}=\frac{\mathrm{T}^{2} \times \mathrm{g}^{2}}{4}\) \(\mathrm{H}=\mathrm{T}^{2}\left(\frac{\mathrm{g}}{8}\right)\) \(\frac{\mathrm{g}}{8} \rightarrow \text { constant }\) So, \(\frac{\Delta H}{H}=\frac{2 \Delta T}{T}\) \(\frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta H}{H}\) \(\frac{\Delta T}{T} \times 100=\frac{1}{2}\left(\frac{\Delta H}{H} \times 100\right) \quad\left(\because \frac{\Delta H}{H} \times 100=10\right)\) \(\frac{\Delta T}{T} \times 100=\frac{1}{2} \times 10\) \(\frac{\Delta T}{T} \times 100=5\) \(\%\) increases in \(\mathrm{T}=5 \%\)
143805
From the top of a tower \(19.6 \mathrm{~m}\) high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of \(45^{\circ}\) with the horizontal, then the time initial velocity of the ball is
1 \(9.8 \mathrm{~ms}^{-1}\)
2 \(4.9 \mathrm{~ms}^{-1}\)
3 \(14.7 \mathrm{~ms}^{-1}\)
4 \(2.8 \mathrm{~ms}^{-1}\)
Explanation:
A Given, \(\theta=45^{\circ}, \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}\) Top of tower height \((\mathrm{H})=19.6 \mathrm{~m}\) \(\because \quad \theta=45^{\circ}\) \(R=19.6\) and \(\quad \tan \theta=\frac{H}{R}\) \(\tan 45^{\circ}=\frac{H}{R}\) \(R=H=19.6 \mathrm{~m}\) In Horizontal direction \(\mathrm{R}=19.6=\mathrm{u} \times \mathrm{t}\) In vertical direction \(\mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\) \(\mathrm{t}=\sqrt{\frac{2 \times 19.6}{9.8}}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \sec\) By putting value of \(t\) in equation (1), we get- \(19.6=\mathrm{u} \times \mathrm{t}\) \(19.6=\mathrm{u} \times 2\) \(\mathrm{u}=\frac{19.6}{2}\) \(\mathrm{u}=9.8 \mathrm{~m} / \mathrm{s}\)
MHT-CET 2020
Motion in Plane
143806
A ball is thrown at two different angles with the same speed \(v\) and from the same point and it has the same range in both the cases. If \(y_{1}\) and \(y_{2}\) be the heights attained in the two cases, then \(y_{1}+y_{2}=\ldots\) :
1 \(\mathrm{u}^{2} / \mathrm{g}\)
2 \(2 u^{2} / g\)
3 \(\mathrm{u}^{2} / 2 \mathrm{~g}\)
4 \(\mathrm{u}^{2} / 4 \mathrm{~g}\)
Explanation:
C Same range can be obtained from complementary angles i.e. \(\theta\) and \(\left(90^{\circ}-\theta\right)\) Height attained by \(\mathrm{I}^{\text {st }}\) ball \(\mathrm{y}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Height attained by II \(^{\text {nd }}\) ball \(\mathrm{y}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}(90-\theta)}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}}\) \(\therefore \quad y_{1}+y_{2}=\frac{u^{2} \sin ^{2} \theta}{2 g}+\frac{u^{2} \cos ^{2} \theta}{2 g}\) \(=\frac{u^{2} \sin ^{2} \theta+u^{2} \cos ^{2} \theta}{2 g}\) \(=\frac{u^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)}{2 g} \quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]\) \(=\frac{u^{2}}{2 g}\)
MHT-CET 2020
Motion in Plane
143807
At what angle with the horizontal should a ball be thrown so that its range \(R\) is related to the time of flight as \(R=\mathbf{5} T^{2}\). (Take \(g=10 \mathrm{~ms}^{-2}\) )
143808
The maximum height attained by a projectile is increased by \(10 \%\). Keeping the angle of projection constant, what is percentage increase in the time of flight?
1 \(5 \%\)
2 \(10 \%\)
3 \(20 \%\)
4 \(40 \%\)
Explanation:
A We know, \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Time of flight \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\). Squaring on both side of equation (2), we get- \(\mathrm{T}^{2}=\left(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\right)^{2}=\frac{4 \mathrm{u}^{2} \sin ^{2} \theta}{\mathrm{g}^{2}}\) \(\mathrm{u}^{2} \sin ^{2} \theta=\frac{\mathrm{T}^{2} \times \mathrm{g}^{2}}{4}\) Putting the value of \(u^{2} \sin ^{2} \theta\) from equation (1), we get- \(\mathrm{H} \times 2 \mathrm{~g}=\frac{\mathrm{T}^{2} \times \mathrm{g}^{2}}{4}\) \(\mathrm{H}=\mathrm{T}^{2}\left(\frac{\mathrm{g}}{8}\right)\) \(\frac{\mathrm{g}}{8} \rightarrow \text { constant }\) So, \(\frac{\Delta H}{H}=\frac{2 \Delta T}{T}\) \(\frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta H}{H}\) \(\frac{\Delta T}{T} \times 100=\frac{1}{2}\left(\frac{\Delta H}{H} \times 100\right) \quad\left(\because \frac{\Delta H}{H} \times 100=10\right)\) \(\frac{\Delta T}{T} \times 100=\frac{1}{2} \times 10\) \(\frac{\Delta T}{T} \times 100=5\) \(\%\) increases in \(\mathrm{T}=5 \%\)
143805
From the top of a tower \(19.6 \mathrm{~m}\) high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of \(45^{\circ}\) with the horizontal, then the time initial velocity of the ball is
1 \(9.8 \mathrm{~ms}^{-1}\)
2 \(4.9 \mathrm{~ms}^{-1}\)
3 \(14.7 \mathrm{~ms}^{-1}\)
4 \(2.8 \mathrm{~ms}^{-1}\)
Explanation:
A Given, \(\theta=45^{\circ}, \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}\) Top of tower height \((\mathrm{H})=19.6 \mathrm{~m}\) \(\because \quad \theta=45^{\circ}\) \(R=19.6\) and \(\quad \tan \theta=\frac{H}{R}\) \(\tan 45^{\circ}=\frac{H}{R}\) \(R=H=19.6 \mathrm{~m}\) In Horizontal direction \(\mathrm{R}=19.6=\mathrm{u} \times \mathrm{t}\) In vertical direction \(\mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\) \(\mathrm{t}=\sqrt{\frac{2 \times 19.6}{9.8}}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \sec\) By putting value of \(t\) in equation (1), we get- \(19.6=\mathrm{u} \times \mathrm{t}\) \(19.6=\mathrm{u} \times 2\) \(\mathrm{u}=\frac{19.6}{2}\) \(\mathrm{u}=9.8 \mathrm{~m} / \mathrm{s}\)
MHT-CET 2020
Motion in Plane
143806
A ball is thrown at two different angles with the same speed \(v\) and from the same point and it has the same range in both the cases. If \(y_{1}\) and \(y_{2}\) be the heights attained in the two cases, then \(y_{1}+y_{2}=\ldots\) :
1 \(\mathrm{u}^{2} / \mathrm{g}\)
2 \(2 u^{2} / g\)
3 \(\mathrm{u}^{2} / 2 \mathrm{~g}\)
4 \(\mathrm{u}^{2} / 4 \mathrm{~g}\)
Explanation:
C Same range can be obtained from complementary angles i.e. \(\theta\) and \(\left(90^{\circ}-\theta\right)\) Height attained by \(\mathrm{I}^{\text {st }}\) ball \(\mathrm{y}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Height attained by II \(^{\text {nd }}\) ball \(\mathrm{y}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}(90-\theta)}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}}\) \(\therefore \quad y_{1}+y_{2}=\frac{u^{2} \sin ^{2} \theta}{2 g}+\frac{u^{2} \cos ^{2} \theta}{2 g}\) \(=\frac{u^{2} \sin ^{2} \theta+u^{2} \cos ^{2} \theta}{2 g}\) \(=\frac{u^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)}{2 g} \quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]\) \(=\frac{u^{2}}{2 g}\)
MHT-CET 2020
Motion in Plane
143807
At what angle with the horizontal should a ball be thrown so that its range \(R\) is related to the time of flight as \(R=\mathbf{5} T^{2}\). (Take \(g=10 \mathrm{~ms}^{-2}\) )
143808
The maximum height attained by a projectile is increased by \(10 \%\). Keeping the angle of projection constant, what is percentage increase in the time of flight?
1 \(5 \%\)
2 \(10 \%\)
3 \(20 \%\)
4 \(40 \%\)
Explanation:
A We know, \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Time of flight \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\). Squaring on both side of equation (2), we get- \(\mathrm{T}^{2}=\left(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\right)^{2}=\frac{4 \mathrm{u}^{2} \sin ^{2} \theta}{\mathrm{g}^{2}}\) \(\mathrm{u}^{2} \sin ^{2} \theta=\frac{\mathrm{T}^{2} \times \mathrm{g}^{2}}{4}\) Putting the value of \(u^{2} \sin ^{2} \theta\) from equation (1), we get- \(\mathrm{H} \times 2 \mathrm{~g}=\frac{\mathrm{T}^{2} \times \mathrm{g}^{2}}{4}\) \(\mathrm{H}=\mathrm{T}^{2}\left(\frac{\mathrm{g}}{8}\right)\) \(\frac{\mathrm{g}}{8} \rightarrow \text { constant }\) So, \(\frac{\Delta H}{H}=\frac{2 \Delta T}{T}\) \(\frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta H}{H}\) \(\frac{\Delta T}{T} \times 100=\frac{1}{2}\left(\frac{\Delta H}{H} \times 100\right) \quad\left(\because \frac{\Delta H}{H} \times 100=10\right)\) \(\frac{\Delta T}{T} \times 100=\frac{1}{2} \times 10\) \(\frac{\Delta T}{T} \times 100=5\) \(\%\) increases in \(\mathrm{T}=5 \%\)
