362375
A ball is thrown upwards with a speed \(u\) from a height \(h\) above the ground. The time taken by the ball to hit the ground is
1 \(\sqrt{2 h / g}\)
2 \(\sqrt{8 h / g}\)
3 \(\dfrac{\sqrt{u^{2}+2 g h}}{g}\)
4 \(\dfrac{u}{g}+\sqrt{\dfrac{2 h}{g}}\)
Explanation:
Time taken to reach the highest point from the height \(h\) is obtained from equation, \(v = u - gt\) \(\therefore \quad 0 = u - gt\,\,{\rm{or}}\,\,t = \frac{u}{g}\) Height attained above \(h\) is obtained from \({v^2} - {u^2} = 2( - g){h_1}\) or \(0 - {u^2} = 2( - g){h_1}\) or \({h_1} = \frac{{{u^2}}}{{2g}}\) Total height, \({h_2} = \left[ {{h_1} + h = \frac{{{u^2}}}{{2g}} + h} \right]\) Time taken to hit the ground is obtained from \(\quad {h_2} = ut + \frac{1}{2}a{t^2}\) or\(\quad \frac{{{u^2}}}{{2g}} + h = 0 + \frac{1}{2}g{t^2}\) \(\therefore \)Total time taken \(t = \sqrt {\frac{{\left( {{u^2} + 2gh} \right)}}{g}} \)
PHXI03:MOTION IN A STRAIGHT LINE
362376
A balloon starts rising from the ground with an acceleration of \(1.25\;m{s^{ - 2}}\). After \(8\;s\), a stone is released from the balloon. The stone will (Taking \(g = 10\;m{s^{ - 2}}\))
1 begin to move down after being released
2 reach the ground in \(4\;s\)
3 cover a distance of \(40\;m\) in reaching the ground
362377
A stone falls freely under gravity. It covers distance \({h_1},{h_2}\,{\rm{and}}\,{h_3}\) in the first \(5 \, s\), the next \(5 \, s\) and the next \(5 \, s\) respectively. The relation between \({h_1},{h_2}\,{\rm{and}}\,{h_3}\) is
For free fall from a height, \(u = 0\) \(\therefore \) Distance covered by stone in first \(5 \, s\), \({h_1} = 0 + \frac{1}{2}\,g{(5)^2} = \frac{{25}}{2}g\) (1) \(\therefore \) Distance covered in first \(10 \, s\), \({s_2} = 0 + \frac{1}{2}g{(10)^2} = \frac{{100}}{2}g\) \(\therefore \) Distance covered in second \(5 \, s\) \({h_2} = {s_2} - {h_1} = \frac{{100}}{2}g - \frac{{25}}{2}g = \frac{{75}}{2}g\) (2) Distance covered in first \(15 \, s\), \({s_3} = 0 + \frac{1}{2}g\;{(15)^2} = \frac{{225}}{2}g\) \(\therefore \) Distance covered in last \(5 \,s\), \({h_3} = {s_3} - {s_2} = \frac{{225}}{2}g - \frac{{100}}{2}g = \frac{{125}}{2}g\) (3) From Eqs (1), (2) and (3), we get \({h_1}:{h_2}:{h_3} = \frac{{25}}{2}g:\frac{{75}}{2}g:\frac{{125}}{2}g = 1:3:5\) \( \Rightarrow \quad {h_1} = \frac{{{h_2}}}{3} = \frac{{{h_3}}}{5}\)
PHXI03:MOTION IN A STRAIGHT LINE
362378
A balloon is at a height of 81 \(m\) and is ascending upwards with a velocity of \(12\,m/s\). A body of \(2 \, kg\) weight is dropped from it. If \(g = 10\,m/{\sec ^2},\) the body will reach the surface of the earth in
362375
A ball is thrown upwards with a speed \(u\) from a height \(h\) above the ground. The time taken by the ball to hit the ground is
1 \(\sqrt{2 h / g}\)
2 \(\sqrt{8 h / g}\)
3 \(\dfrac{\sqrt{u^{2}+2 g h}}{g}\)
4 \(\dfrac{u}{g}+\sqrt{\dfrac{2 h}{g}}\)
Explanation:
Time taken to reach the highest point from the height \(h\) is obtained from equation, \(v = u - gt\) \(\therefore \quad 0 = u - gt\,\,{\rm{or}}\,\,t = \frac{u}{g}\) Height attained above \(h\) is obtained from \({v^2} - {u^2} = 2( - g){h_1}\) or \(0 - {u^2} = 2( - g){h_1}\) or \({h_1} = \frac{{{u^2}}}{{2g}}\) Total height, \({h_2} = \left[ {{h_1} + h = \frac{{{u^2}}}{{2g}} + h} \right]\) Time taken to hit the ground is obtained from \(\quad {h_2} = ut + \frac{1}{2}a{t^2}\) or\(\quad \frac{{{u^2}}}{{2g}} + h = 0 + \frac{1}{2}g{t^2}\) \(\therefore \)Total time taken \(t = \sqrt {\frac{{\left( {{u^2} + 2gh} \right)}}{g}} \)
PHXI03:MOTION IN A STRAIGHT LINE
362376
A balloon starts rising from the ground with an acceleration of \(1.25\;m{s^{ - 2}}\). After \(8\;s\), a stone is released from the balloon. The stone will (Taking \(g = 10\;m{s^{ - 2}}\))
1 begin to move down after being released
2 reach the ground in \(4\;s\)
3 cover a distance of \(40\;m\) in reaching the ground
362377
A stone falls freely under gravity. It covers distance \({h_1},{h_2}\,{\rm{and}}\,{h_3}\) in the first \(5 \, s\), the next \(5 \, s\) and the next \(5 \, s\) respectively. The relation between \({h_1},{h_2}\,{\rm{and}}\,{h_3}\) is
For free fall from a height, \(u = 0\) \(\therefore \) Distance covered by stone in first \(5 \, s\), \({h_1} = 0 + \frac{1}{2}\,g{(5)^2} = \frac{{25}}{2}g\) (1) \(\therefore \) Distance covered in first \(10 \, s\), \({s_2} = 0 + \frac{1}{2}g{(10)^2} = \frac{{100}}{2}g\) \(\therefore \) Distance covered in second \(5 \, s\) \({h_2} = {s_2} - {h_1} = \frac{{100}}{2}g - \frac{{25}}{2}g = \frac{{75}}{2}g\) (2) Distance covered in first \(15 \, s\), \({s_3} = 0 + \frac{1}{2}g\;{(15)^2} = \frac{{225}}{2}g\) \(\therefore \) Distance covered in last \(5 \,s\), \({h_3} = {s_3} - {s_2} = \frac{{225}}{2}g - \frac{{100}}{2}g = \frac{{125}}{2}g\) (3) From Eqs (1), (2) and (3), we get \({h_1}:{h_2}:{h_3} = \frac{{25}}{2}g:\frac{{75}}{2}g:\frac{{125}}{2}g = 1:3:5\) \( \Rightarrow \quad {h_1} = \frac{{{h_2}}}{3} = \frac{{{h_3}}}{5}\)
PHXI03:MOTION IN A STRAIGHT LINE
362378
A balloon is at a height of 81 \(m\) and is ascending upwards with a velocity of \(12\,m/s\). A body of \(2 \, kg\) weight is dropped from it. If \(g = 10\,m/{\sec ^2},\) the body will reach the surface of the earth in
362375
A ball is thrown upwards with a speed \(u\) from a height \(h\) above the ground. The time taken by the ball to hit the ground is
1 \(\sqrt{2 h / g}\)
2 \(\sqrt{8 h / g}\)
3 \(\dfrac{\sqrt{u^{2}+2 g h}}{g}\)
4 \(\dfrac{u}{g}+\sqrt{\dfrac{2 h}{g}}\)
Explanation:
Time taken to reach the highest point from the height \(h\) is obtained from equation, \(v = u - gt\) \(\therefore \quad 0 = u - gt\,\,{\rm{or}}\,\,t = \frac{u}{g}\) Height attained above \(h\) is obtained from \({v^2} - {u^2} = 2( - g){h_1}\) or \(0 - {u^2} = 2( - g){h_1}\) or \({h_1} = \frac{{{u^2}}}{{2g}}\) Total height, \({h_2} = \left[ {{h_1} + h = \frac{{{u^2}}}{{2g}} + h} \right]\) Time taken to hit the ground is obtained from \(\quad {h_2} = ut + \frac{1}{2}a{t^2}\) or\(\quad \frac{{{u^2}}}{{2g}} + h = 0 + \frac{1}{2}g{t^2}\) \(\therefore \)Total time taken \(t = \sqrt {\frac{{\left( {{u^2} + 2gh} \right)}}{g}} \)
PHXI03:MOTION IN A STRAIGHT LINE
362376
A balloon starts rising from the ground with an acceleration of \(1.25\;m{s^{ - 2}}\). After \(8\;s\), a stone is released from the balloon. The stone will (Taking \(g = 10\;m{s^{ - 2}}\))
1 begin to move down after being released
2 reach the ground in \(4\;s\)
3 cover a distance of \(40\;m\) in reaching the ground
362377
A stone falls freely under gravity. It covers distance \({h_1},{h_2}\,{\rm{and}}\,{h_3}\) in the first \(5 \, s\), the next \(5 \, s\) and the next \(5 \, s\) respectively. The relation between \({h_1},{h_2}\,{\rm{and}}\,{h_3}\) is
For free fall from a height, \(u = 0\) \(\therefore \) Distance covered by stone in first \(5 \, s\), \({h_1} = 0 + \frac{1}{2}\,g{(5)^2} = \frac{{25}}{2}g\) (1) \(\therefore \) Distance covered in first \(10 \, s\), \({s_2} = 0 + \frac{1}{2}g{(10)^2} = \frac{{100}}{2}g\) \(\therefore \) Distance covered in second \(5 \, s\) \({h_2} = {s_2} - {h_1} = \frac{{100}}{2}g - \frac{{25}}{2}g = \frac{{75}}{2}g\) (2) Distance covered in first \(15 \, s\), \({s_3} = 0 + \frac{1}{2}g\;{(15)^2} = \frac{{225}}{2}g\) \(\therefore \) Distance covered in last \(5 \,s\), \({h_3} = {s_3} - {s_2} = \frac{{225}}{2}g - \frac{{100}}{2}g = \frac{{125}}{2}g\) (3) From Eqs (1), (2) and (3), we get \({h_1}:{h_2}:{h_3} = \frac{{25}}{2}g:\frac{{75}}{2}g:\frac{{125}}{2}g = 1:3:5\) \( \Rightarrow \quad {h_1} = \frac{{{h_2}}}{3} = \frac{{{h_3}}}{5}\)
PHXI03:MOTION IN A STRAIGHT LINE
362378
A balloon is at a height of 81 \(m\) and is ascending upwards with a velocity of \(12\,m/s\). A body of \(2 \, kg\) weight is dropped from it. If \(g = 10\,m/{\sec ^2},\) the body will reach the surface of the earth in
362375
A ball is thrown upwards with a speed \(u\) from a height \(h\) above the ground. The time taken by the ball to hit the ground is
1 \(\sqrt{2 h / g}\)
2 \(\sqrt{8 h / g}\)
3 \(\dfrac{\sqrt{u^{2}+2 g h}}{g}\)
4 \(\dfrac{u}{g}+\sqrt{\dfrac{2 h}{g}}\)
Explanation:
Time taken to reach the highest point from the height \(h\) is obtained from equation, \(v = u - gt\) \(\therefore \quad 0 = u - gt\,\,{\rm{or}}\,\,t = \frac{u}{g}\) Height attained above \(h\) is obtained from \({v^2} - {u^2} = 2( - g){h_1}\) or \(0 - {u^2} = 2( - g){h_1}\) or \({h_1} = \frac{{{u^2}}}{{2g}}\) Total height, \({h_2} = \left[ {{h_1} + h = \frac{{{u^2}}}{{2g}} + h} \right]\) Time taken to hit the ground is obtained from \(\quad {h_2} = ut + \frac{1}{2}a{t^2}\) or\(\quad \frac{{{u^2}}}{{2g}} + h = 0 + \frac{1}{2}g{t^2}\) \(\therefore \)Total time taken \(t = \sqrt {\frac{{\left( {{u^2} + 2gh} \right)}}{g}} \)
PHXI03:MOTION IN A STRAIGHT LINE
362376
A balloon starts rising from the ground with an acceleration of \(1.25\;m{s^{ - 2}}\). After \(8\;s\), a stone is released from the balloon. The stone will (Taking \(g = 10\;m{s^{ - 2}}\))
1 begin to move down after being released
2 reach the ground in \(4\;s\)
3 cover a distance of \(40\;m\) in reaching the ground
362377
A stone falls freely under gravity. It covers distance \({h_1},{h_2}\,{\rm{and}}\,{h_3}\) in the first \(5 \, s\), the next \(5 \, s\) and the next \(5 \, s\) respectively. The relation between \({h_1},{h_2}\,{\rm{and}}\,{h_3}\) is
For free fall from a height, \(u = 0\) \(\therefore \) Distance covered by stone in first \(5 \, s\), \({h_1} = 0 + \frac{1}{2}\,g{(5)^2} = \frac{{25}}{2}g\) (1) \(\therefore \) Distance covered in first \(10 \, s\), \({s_2} = 0 + \frac{1}{2}g{(10)^2} = \frac{{100}}{2}g\) \(\therefore \) Distance covered in second \(5 \, s\) \({h_2} = {s_2} - {h_1} = \frac{{100}}{2}g - \frac{{25}}{2}g = \frac{{75}}{2}g\) (2) Distance covered in first \(15 \, s\), \({s_3} = 0 + \frac{1}{2}g\;{(15)^2} = \frac{{225}}{2}g\) \(\therefore \) Distance covered in last \(5 \,s\), \({h_3} = {s_3} - {s_2} = \frac{{225}}{2}g - \frac{{100}}{2}g = \frac{{125}}{2}g\) (3) From Eqs (1), (2) and (3), we get \({h_1}:{h_2}:{h_3} = \frac{{25}}{2}g:\frac{{75}}{2}g:\frac{{125}}{2}g = 1:3:5\) \( \Rightarrow \quad {h_1} = \frac{{{h_2}}}{3} = \frac{{{h_3}}}{5}\)
PHXI03:MOTION IN A STRAIGHT LINE
362378
A balloon is at a height of 81 \(m\) and is ascending upwards with a velocity of \(12\,m/s\). A body of \(2 \, kg\) weight is dropped from it. If \(g = 10\,m/{\sec ^2},\) the body will reach the surface of the earth in
