362384
When a ball is thrown up vertically with a velocity \({v_{0}}\) it reaches a height \({h}\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \({\sqrt{3} v_{0}}\)
2 \({3 v_{0}}\)
3 \({9 v_{0}}\)
4 \({3 / 2 v_{0}}\)
Explanation:
Here initial velocity, \({u=v_{0}}\) Height reached \({=h}\) At highest point final velocity \((v)\) will be zero. From equation of motion, \({v^{2}-u^{2}=2 a s}\) \({0-v_{0}^{2}=-2 g h \Rightarrow h=\dfrac{v_{0}^{2}}{2 g}}\) For thrice the maximum height (3h), \(0 - v_0^{\prime 2} = - 2g\left( {\frac{{3v_0^2}}{{2g}}} \right)\) (\(v_0^1\) new initial velocity) \( \Rightarrow v_0^{\prime 2}{\rm{ = 3}}v_0^2 \Rightarrow {{v'}_0} = \sqrt 3 {v_0}\). So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362385
A body is projected downward with speed \(20\,m/s\) from the top of a tower of height 50 \(m\). Find the time at which it hits the ground.
1 \(\sqrt {14} - 2\sec \)
2 \(\sqrt {10} - 2\sec \)
3 \(\sqrt {14} + 2\sec \)
4 \(\sqrt {10} + 2\sec \)
Explanation:
Method-1 Consider the origin at the top of the tower and downward is positive and upward is negative \({S_0} = 0\) \(u = 20\,m/s\) (Going towards \(\left( + \right)ve\) direction) \(a = g\) (Accelerated motion) \(s = 20t + \frac{1}{2}g{t^2}\) When it hits the ground \(s = 50\,m\) \(50 = 20t + \frac{1}{2}10{t^2}\) \({t^2} + 4t - 10 = 0\) \(t = \frac{{ - 4 \pm \sqrt {56} }}{2} = \sqrt {14} - 2\) Method 2 If we do the problem graphically we can avoid quadratic equation. When big numbers are involved solving quadratic equation is difficult. The graph between \(v\) and \(t\) is a straight line since \(a = \frac{{dv}}{{dt}} = g = {\rm{constant}}\) The velocity when the body touches the ground is \({\rm{v}}_{\max }^2 = {20^2} + 2 \times 10 \times 50 \Rightarrow {v_{\max }} = 10\sqrt {14} \) Total distance travelled = Area of the graph \(50 = 20 \times t + \frac{1}{2} \times t \times \left( {10\sqrt {14} - 20} \right)\) \( \Rightarrow t = \frac{{10}}{{2 + \sqrt {14} }}\) on rationalising we get \(t = \sqrt {14} - 2\)
PHXI03:MOTION IN A STRAIGHT LINE
362386
When a ball is thrown up vertically with velocity \(v_{0}\), it reaches a maximum height of \(h\). If one wishes to triple the maximum height, then the ball should be thrown with velocity
1 \(\sqrt 3 \,{v_0}\)
2 \(3 v_{0}\)
3 \(9 v_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
As, \(v^{2}=u^{2}-2 g h \Rightarrow u^{2} \propto h\) Hence, \(\dfrac{u_{1}}{u_{2}}=\sqrt{\dfrac{h_{1}}{h_{2}}}\) \(\Rightarrow \dfrac{v_{0}}{u_{2}}=\sqrt{\dfrac{h}{3 h}}\) (given) \(\Rightarrow u_{2}=\sqrt{3} v_{0}\)
AIIMS - 2005
PHXI03:MOTION IN A STRAIGHT LINE
362387
A ball is thrown vertically downward with a velocity of \(20\,m/s\) from the top of a tower. It hits the ground after some time with a velocity of \(80\,m/s\).The height of the tower is: \((g = 10\;m/{s^2})\)
362384
When a ball is thrown up vertically with a velocity \({v_{0}}\) it reaches a height \({h}\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \({\sqrt{3} v_{0}}\)
2 \({3 v_{0}}\)
3 \({9 v_{0}}\)
4 \({3 / 2 v_{0}}\)
Explanation:
Here initial velocity, \({u=v_{0}}\) Height reached \({=h}\) At highest point final velocity \((v)\) will be zero. From equation of motion, \({v^{2}-u^{2}=2 a s}\) \({0-v_{0}^{2}=-2 g h \Rightarrow h=\dfrac{v_{0}^{2}}{2 g}}\) For thrice the maximum height (3h), \(0 - v_0^{\prime 2} = - 2g\left( {\frac{{3v_0^2}}{{2g}}} \right)\) (\(v_0^1\) new initial velocity) \( \Rightarrow v_0^{\prime 2}{\rm{ = 3}}v_0^2 \Rightarrow {{v'}_0} = \sqrt 3 {v_0}\). So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362385
A body is projected downward with speed \(20\,m/s\) from the top of a tower of height 50 \(m\). Find the time at which it hits the ground.
1 \(\sqrt {14} - 2\sec \)
2 \(\sqrt {10} - 2\sec \)
3 \(\sqrt {14} + 2\sec \)
4 \(\sqrt {10} + 2\sec \)
Explanation:
Method-1 Consider the origin at the top of the tower and downward is positive and upward is negative \({S_0} = 0\) \(u = 20\,m/s\) (Going towards \(\left( + \right)ve\) direction) \(a = g\) (Accelerated motion) \(s = 20t + \frac{1}{2}g{t^2}\) When it hits the ground \(s = 50\,m\) \(50 = 20t + \frac{1}{2}10{t^2}\) \({t^2} + 4t - 10 = 0\) \(t = \frac{{ - 4 \pm \sqrt {56} }}{2} = \sqrt {14} - 2\) Method 2 If we do the problem graphically we can avoid quadratic equation. When big numbers are involved solving quadratic equation is difficult. The graph between \(v\) and \(t\) is a straight line since \(a = \frac{{dv}}{{dt}} = g = {\rm{constant}}\) The velocity when the body touches the ground is \({\rm{v}}_{\max }^2 = {20^2} + 2 \times 10 \times 50 \Rightarrow {v_{\max }} = 10\sqrt {14} \) Total distance travelled = Area of the graph \(50 = 20 \times t + \frac{1}{2} \times t \times \left( {10\sqrt {14} - 20} \right)\) \( \Rightarrow t = \frac{{10}}{{2 + \sqrt {14} }}\) on rationalising we get \(t = \sqrt {14} - 2\)
PHXI03:MOTION IN A STRAIGHT LINE
362386
When a ball is thrown up vertically with velocity \(v_{0}\), it reaches a maximum height of \(h\). If one wishes to triple the maximum height, then the ball should be thrown with velocity
1 \(\sqrt 3 \,{v_0}\)
2 \(3 v_{0}\)
3 \(9 v_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
As, \(v^{2}=u^{2}-2 g h \Rightarrow u^{2} \propto h\) Hence, \(\dfrac{u_{1}}{u_{2}}=\sqrt{\dfrac{h_{1}}{h_{2}}}\) \(\Rightarrow \dfrac{v_{0}}{u_{2}}=\sqrt{\dfrac{h}{3 h}}\) (given) \(\Rightarrow u_{2}=\sqrt{3} v_{0}\)
AIIMS - 2005
PHXI03:MOTION IN A STRAIGHT LINE
362387
A ball is thrown vertically downward with a velocity of \(20\,m/s\) from the top of a tower. It hits the ground after some time with a velocity of \(80\,m/s\).The height of the tower is: \((g = 10\;m/{s^2})\)
362384
When a ball is thrown up vertically with a velocity \({v_{0}}\) it reaches a height \({h}\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \({\sqrt{3} v_{0}}\)
2 \({3 v_{0}}\)
3 \({9 v_{0}}\)
4 \({3 / 2 v_{0}}\)
Explanation:
Here initial velocity, \({u=v_{0}}\) Height reached \({=h}\) At highest point final velocity \((v)\) will be zero. From equation of motion, \({v^{2}-u^{2}=2 a s}\) \({0-v_{0}^{2}=-2 g h \Rightarrow h=\dfrac{v_{0}^{2}}{2 g}}\) For thrice the maximum height (3h), \(0 - v_0^{\prime 2} = - 2g\left( {\frac{{3v_0^2}}{{2g}}} \right)\) (\(v_0^1\) new initial velocity) \( \Rightarrow v_0^{\prime 2}{\rm{ = 3}}v_0^2 \Rightarrow {{v'}_0} = \sqrt 3 {v_0}\). So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362385
A body is projected downward with speed \(20\,m/s\) from the top of a tower of height 50 \(m\). Find the time at which it hits the ground.
1 \(\sqrt {14} - 2\sec \)
2 \(\sqrt {10} - 2\sec \)
3 \(\sqrt {14} + 2\sec \)
4 \(\sqrt {10} + 2\sec \)
Explanation:
Method-1 Consider the origin at the top of the tower and downward is positive and upward is negative \({S_0} = 0\) \(u = 20\,m/s\) (Going towards \(\left( + \right)ve\) direction) \(a = g\) (Accelerated motion) \(s = 20t + \frac{1}{2}g{t^2}\) When it hits the ground \(s = 50\,m\) \(50 = 20t + \frac{1}{2}10{t^2}\) \({t^2} + 4t - 10 = 0\) \(t = \frac{{ - 4 \pm \sqrt {56} }}{2} = \sqrt {14} - 2\) Method 2 If we do the problem graphically we can avoid quadratic equation. When big numbers are involved solving quadratic equation is difficult. The graph between \(v\) and \(t\) is a straight line since \(a = \frac{{dv}}{{dt}} = g = {\rm{constant}}\) The velocity when the body touches the ground is \({\rm{v}}_{\max }^2 = {20^2} + 2 \times 10 \times 50 \Rightarrow {v_{\max }} = 10\sqrt {14} \) Total distance travelled = Area of the graph \(50 = 20 \times t + \frac{1}{2} \times t \times \left( {10\sqrt {14} - 20} \right)\) \( \Rightarrow t = \frac{{10}}{{2 + \sqrt {14} }}\) on rationalising we get \(t = \sqrt {14} - 2\)
PHXI03:MOTION IN A STRAIGHT LINE
362386
When a ball is thrown up vertically with velocity \(v_{0}\), it reaches a maximum height of \(h\). If one wishes to triple the maximum height, then the ball should be thrown with velocity
1 \(\sqrt 3 \,{v_0}\)
2 \(3 v_{0}\)
3 \(9 v_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
As, \(v^{2}=u^{2}-2 g h \Rightarrow u^{2} \propto h\) Hence, \(\dfrac{u_{1}}{u_{2}}=\sqrt{\dfrac{h_{1}}{h_{2}}}\) \(\Rightarrow \dfrac{v_{0}}{u_{2}}=\sqrt{\dfrac{h}{3 h}}\) (given) \(\Rightarrow u_{2}=\sqrt{3} v_{0}\)
AIIMS - 2005
PHXI03:MOTION IN A STRAIGHT LINE
362387
A ball is thrown vertically downward with a velocity of \(20\,m/s\) from the top of a tower. It hits the ground after some time with a velocity of \(80\,m/s\).The height of the tower is: \((g = 10\;m/{s^2})\)
362384
When a ball is thrown up vertically with a velocity \({v_{0}}\) it reaches a height \({h}\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \({\sqrt{3} v_{0}}\)
2 \({3 v_{0}}\)
3 \({9 v_{0}}\)
4 \({3 / 2 v_{0}}\)
Explanation:
Here initial velocity, \({u=v_{0}}\) Height reached \({=h}\) At highest point final velocity \((v)\) will be zero. From equation of motion, \({v^{2}-u^{2}=2 a s}\) \({0-v_{0}^{2}=-2 g h \Rightarrow h=\dfrac{v_{0}^{2}}{2 g}}\) For thrice the maximum height (3h), \(0 - v_0^{\prime 2} = - 2g\left( {\frac{{3v_0^2}}{{2g}}} \right)\) (\(v_0^1\) new initial velocity) \( \Rightarrow v_0^{\prime 2}{\rm{ = 3}}v_0^2 \Rightarrow {{v'}_0} = \sqrt 3 {v_0}\). So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362385
A body is projected downward with speed \(20\,m/s\) from the top of a tower of height 50 \(m\). Find the time at which it hits the ground.
1 \(\sqrt {14} - 2\sec \)
2 \(\sqrt {10} - 2\sec \)
3 \(\sqrt {14} + 2\sec \)
4 \(\sqrt {10} + 2\sec \)
Explanation:
Method-1 Consider the origin at the top of the tower and downward is positive and upward is negative \({S_0} = 0\) \(u = 20\,m/s\) (Going towards \(\left( + \right)ve\) direction) \(a = g\) (Accelerated motion) \(s = 20t + \frac{1}{2}g{t^2}\) When it hits the ground \(s = 50\,m\) \(50 = 20t + \frac{1}{2}10{t^2}\) \({t^2} + 4t - 10 = 0\) \(t = \frac{{ - 4 \pm \sqrt {56} }}{2} = \sqrt {14} - 2\) Method 2 If we do the problem graphically we can avoid quadratic equation. When big numbers are involved solving quadratic equation is difficult. The graph between \(v\) and \(t\) is a straight line since \(a = \frac{{dv}}{{dt}} = g = {\rm{constant}}\) The velocity when the body touches the ground is \({\rm{v}}_{\max }^2 = {20^2} + 2 \times 10 \times 50 \Rightarrow {v_{\max }} = 10\sqrt {14} \) Total distance travelled = Area of the graph \(50 = 20 \times t + \frac{1}{2} \times t \times \left( {10\sqrt {14} - 20} \right)\) \( \Rightarrow t = \frac{{10}}{{2 + \sqrt {14} }}\) on rationalising we get \(t = \sqrt {14} - 2\)
PHXI03:MOTION IN A STRAIGHT LINE
362386
When a ball is thrown up vertically with velocity \(v_{0}\), it reaches a maximum height of \(h\). If one wishes to triple the maximum height, then the ball should be thrown with velocity
1 \(\sqrt 3 \,{v_0}\)
2 \(3 v_{0}\)
3 \(9 v_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
As, \(v^{2}=u^{2}-2 g h \Rightarrow u^{2} \propto h\) Hence, \(\dfrac{u_{1}}{u_{2}}=\sqrt{\dfrac{h_{1}}{h_{2}}}\) \(\Rightarrow \dfrac{v_{0}}{u_{2}}=\sqrt{\dfrac{h}{3 h}}\) (given) \(\Rightarrow u_{2}=\sqrt{3} v_{0}\)
AIIMS - 2005
PHXI03:MOTION IN A STRAIGHT LINE
362387
A ball is thrown vertically downward with a velocity of \(20\,m/s\) from the top of a tower. It hits the ground after some time with a velocity of \(80\,m/s\).The height of the tower is: \((g = 10\;m/{s^2})\)
362384
When a ball is thrown up vertically with a velocity \({v_{0}}\) it reaches a height \({h}\). If one wishes to triple the maximum height, then the ball should be thrown with a velocity of
1 \({\sqrt{3} v_{0}}\)
2 \({3 v_{0}}\)
3 \({9 v_{0}}\)
4 \({3 / 2 v_{0}}\)
Explanation:
Here initial velocity, \({u=v_{0}}\) Height reached \({=h}\) At highest point final velocity \((v)\) will be zero. From equation of motion, \({v^{2}-u^{2}=2 a s}\) \({0-v_{0}^{2}=-2 g h \Rightarrow h=\dfrac{v_{0}^{2}}{2 g}}\) For thrice the maximum height (3h), \(0 - v_0^{\prime 2} = - 2g\left( {\frac{{3v_0^2}}{{2g}}} \right)\) (\(v_0^1\) new initial velocity) \( \Rightarrow v_0^{\prime 2}{\rm{ = 3}}v_0^2 \Rightarrow {{v'}_0} = \sqrt 3 {v_0}\). So correct option is (1)
PHXI03:MOTION IN A STRAIGHT LINE
362385
A body is projected downward with speed \(20\,m/s\) from the top of a tower of height 50 \(m\). Find the time at which it hits the ground.
1 \(\sqrt {14} - 2\sec \)
2 \(\sqrt {10} - 2\sec \)
3 \(\sqrt {14} + 2\sec \)
4 \(\sqrt {10} + 2\sec \)
Explanation:
Method-1 Consider the origin at the top of the tower and downward is positive and upward is negative \({S_0} = 0\) \(u = 20\,m/s\) (Going towards \(\left( + \right)ve\) direction) \(a = g\) (Accelerated motion) \(s = 20t + \frac{1}{2}g{t^2}\) When it hits the ground \(s = 50\,m\) \(50 = 20t + \frac{1}{2}10{t^2}\) \({t^2} + 4t - 10 = 0\) \(t = \frac{{ - 4 \pm \sqrt {56} }}{2} = \sqrt {14} - 2\) Method 2 If we do the problem graphically we can avoid quadratic equation. When big numbers are involved solving quadratic equation is difficult. The graph between \(v\) and \(t\) is a straight line since \(a = \frac{{dv}}{{dt}} = g = {\rm{constant}}\) The velocity when the body touches the ground is \({\rm{v}}_{\max }^2 = {20^2} + 2 \times 10 \times 50 \Rightarrow {v_{\max }} = 10\sqrt {14} \) Total distance travelled = Area of the graph \(50 = 20 \times t + \frac{1}{2} \times t \times \left( {10\sqrt {14} - 20} \right)\) \( \Rightarrow t = \frac{{10}}{{2 + \sqrt {14} }}\) on rationalising we get \(t = \sqrt {14} - 2\)
PHXI03:MOTION IN A STRAIGHT LINE
362386
When a ball is thrown up vertically with velocity \(v_{0}\), it reaches a maximum height of \(h\). If one wishes to triple the maximum height, then the ball should be thrown with velocity
1 \(\sqrt 3 \,{v_0}\)
2 \(3 v_{0}\)
3 \(9 v_{0}\)
4 \(3 / 2 v_{0}\)
Explanation:
As, \(v^{2}=u^{2}-2 g h \Rightarrow u^{2} \propto h\) Hence, \(\dfrac{u_{1}}{u_{2}}=\sqrt{\dfrac{h_{1}}{h_{2}}}\) \(\Rightarrow \dfrac{v_{0}}{u_{2}}=\sqrt{\dfrac{h}{3 h}}\) (given) \(\Rightarrow u_{2}=\sqrt{3} v_{0}\)
AIIMS - 2005
PHXI03:MOTION IN A STRAIGHT LINE
362387
A ball is thrown vertically downward with a velocity of \(20\,m/s\) from the top of a tower. It hits the ground after some time with a velocity of \(80\,m/s\).The height of the tower is: \((g = 10\;m/{s^2})\)
