362366
A body is projected vertically upwards. The times corresponding to height \(h\) while ascending and while descending are \(t_{1}\) and \(t_{2}\), respectively. Then, the velocity of projection will be (take \(g\) as acceleration due to gravity)
Let \(v\) be initial velocity of vertical projection and \(t\) be the time taken by the body to reach a height \(h\) from ground. Here, \(\quad u = u,a = - g,s = h,t = t\) Using, \(\quad s = ut + \frac{1}{2}a{t^2}\), We have \(h = ut + \frac{1}{2}( - g){t^2}\) or \(g{t^2} - 2ut + 2h = 0\) \(\therefore \quad t = \frac{{2u \pm \sqrt {4{u^2} - 4g \times 2h} }}{{2g}}\) \( = \frac{{u \pm \sqrt {{u^2} - 2gh} }}{g}\) It means \(t\) has two values, i.e. \({t_1} = \frac{{u + \sqrt {{u^2} - 2gh} }}{g}\) \({t_2} = \frac{{u - \sqrt {{u^2} - 2gh} }}{g}\) \({t_1} + {t_2} = \frac{{2u}}{g}\,\,{\rm{or}}\,\,u = \frac{{g\left( {{t_1} + {t_2}} \right)}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362367
A body is dropped from a certain height
1 The time taken to travel first half of the height is less than that for second half
2 The time taken to travel first half of the height is greater than that for second half
3 The time taken to travel first half of the height is equal to that for second half
4 Any one of the above three situations may be correct depending upon the value of height
Explanation:
During first half, velocities are lesser as compared to velocities during second half. So more time is taken in first half.
PHXI03:MOTION IN A STRAIGHT LINE
362368
A small block slides down on a smooth inclined plane, starting from rest at time \(t = 0\). Let \({s_n}\) be the distance travelled by the block in the interval \(t = n - 1\) to \(t = n\). Then , the ratio \(\frac{{{S_n}}}{{{S_{n + 1}}}}\) is
362369
A stone thrown upwards with speed \(u\) attains maximum height \(h\). Another stone thrown upwards from the same point with speed \(2u\) attains maximum height \(H\). What is the relation between \(h\) and \(H\)?
362370
A man throws balls with the same speed vertically upwards one after the other at an interval of \(2 s\). What should be the speed of the throw, so that more than two balls are in the sky at any time? (Take, \(g = 9.8\;m{s^{ - 2}}\) )
1 Any speed less than \(19.6\;m{s^{ - 1}}\)
2 Only with speed \(19.6\;m{s^{ - 1}}\)
3 More than \(19.6\;m{s^{ - 1}}\)
4 Atleast \(9.8\;m{s^{ - 1}}\)
Explanation:
Time taken by ball to reach maximum height, \(v = u - gT\) At maximum height, final speed is zero, i.e. \(v = 0,\,\,\)so,\(u = gT.\) In \(2s,u = 2 \times 9.8 = 19.6\;m{s^{ - 1}}\) If a man throws the ball with velocity of \(19.6\;m{s^{ - 1}},\) then after \(2 s\) it will reach the maximum height. When he throws 2nd ball, 1st is at the top. When he throws third ball, 1 st will come to ground and 2 nd will go at the top. Therefore, only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than \(19.6\;m{s^{ - 1}}\)
362366
A body is projected vertically upwards. The times corresponding to height \(h\) while ascending and while descending are \(t_{1}\) and \(t_{2}\), respectively. Then, the velocity of projection will be (take \(g\) as acceleration due to gravity)
Let \(v\) be initial velocity of vertical projection and \(t\) be the time taken by the body to reach a height \(h\) from ground. Here, \(\quad u = u,a = - g,s = h,t = t\) Using, \(\quad s = ut + \frac{1}{2}a{t^2}\), We have \(h = ut + \frac{1}{2}( - g){t^2}\) or \(g{t^2} - 2ut + 2h = 0\) \(\therefore \quad t = \frac{{2u \pm \sqrt {4{u^2} - 4g \times 2h} }}{{2g}}\) \( = \frac{{u \pm \sqrt {{u^2} - 2gh} }}{g}\) It means \(t\) has two values, i.e. \({t_1} = \frac{{u + \sqrt {{u^2} - 2gh} }}{g}\) \({t_2} = \frac{{u - \sqrt {{u^2} - 2gh} }}{g}\) \({t_1} + {t_2} = \frac{{2u}}{g}\,\,{\rm{or}}\,\,u = \frac{{g\left( {{t_1} + {t_2}} \right)}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362367
A body is dropped from a certain height
1 The time taken to travel first half of the height is less than that for second half
2 The time taken to travel first half of the height is greater than that for second half
3 The time taken to travel first half of the height is equal to that for second half
4 Any one of the above three situations may be correct depending upon the value of height
Explanation:
During first half, velocities are lesser as compared to velocities during second half. So more time is taken in first half.
PHXI03:MOTION IN A STRAIGHT LINE
362368
A small block slides down on a smooth inclined plane, starting from rest at time \(t = 0\). Let \({s_n}\) be the distance travelled by the block in the interval \(t = n - 1\) to \(t = n\). Then , the ratio \(\frac{{{S_n}}}{{{S_{n + 1}}}}\) is
362369
A stone thrown upwards with speed \(u\) attains maximum height \(h\). Another stone thrown upwards from the same point with speed \(2u\) attains maximum height \(H\). What is the relation between \(h\) and \(H\)?
362370
A man throws balls with the same speed vertically upwards one after the other at an interval of \(2 s\). What should be the speed of the throw, so that more than two balls are in the sky at any time? (Take, \(g = 9.8\;m{s^{ - 2}}\) )
1 Any speed less than \(19.6\;m{s^{ - 1}}\)
2 Only with speed \(19.6\;m{s^{ - 1}}\)
3 More than \(19.6\;m{s^{ - 1}}\)
4 Atleast \(9.8\;m{s^{ - 1}}\)
Explanation:
Time taken by ball to reach maximum height, \(v = u - gT\) At maximum height, final speed is zero, i.e. \(v = 0,\,\,\)so,\(u = gT.\) In \(2s,u = 2 \times 9.8 = 19.6\;m{s^{ - 1}}\) If a man throws the ball with velocity of \(19.6\;m{s^{ - 1}},\) then after \(2 s\) it will reach the maximum height. When he throws 2nd ball, 1st is at the top. When he throws third ball, 1 st will come to ground and 2 nd will go at the top. Therefore, only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than \(19.6\;m{s^{ - 1}}\)
362366
A body is projected vertically upwards. The times corresponding to height \(h\) while ascending and while descending are \(t_{1}\) and \(t_{2}\), respectively. Then, the velocity of projection will be (take \(g\) as acceleration due to gravity)
Let \(v\) be initial velocity of vertical projection and \(t\) be the time taken by the body to reach a height \(h\) from ground. Here, \(\quad u = u,a = - g,s = h,t = t\) Using, \(\quad s = ut + \frac{1}{2}a{t^2}\), We have \(h = ut + \frac{1}{2}( - g){t^2}\) or \(g{t^2} - 2ut + 2h = 0\) \(\therefore \quad t = \frac{{2u \pm \sqrt {4{u^2} - 4g \times 2h} }}{{2g}}\) \( = \frac{{u \pm \sqrt {{u^2} - 2gh} }}{g}\) It means \(t\) has two values, i.e. \({t_1} = \frac{{u + \sqrt {{u^2} - 2gh} }}{g}\) \({t_2} = \frac{{u - \sqrt {{u^2} - 2gh} }}{g}\) \({t_1} + {t_2} = \frac{{2u}}{g}\,\,{\rm{or}}\,\,u = \frac{{g\left( {{t_1} + {t_2}} \right)}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362367
A body is dropped from a certain height
1 The time taken to travel first half of the height is less than that for second half
2 The time taken to travel first half of the height is greater than that for second half
3 The time taken to travel first half of the height is equal to that for second half
4 Any one of the above three situations may be correct depending upon the value of height
Explanation:
During first half, velocities are lesser as compared to velocities during second half. So more time is taken in first half.
PHXI03:MOTION IN A STRAIGHT LINE
362368
A small block slides down on a smooth inclined plane, starting from rest at time \(t = 0\). Let \({s_n}\) be the distance travelled by the block in the interval \(t = n - 1\) to \(t = n\). Then , the ratio \(\frac{{{S_n}}}{{{S_{n + 1}}}}\) is
362369
A stone thrown upwards with speed \(u\) attains maximum height \(h\). Another stone thrown upwards from the same point with speed \(2u\) attains maximum height \(H\). What is the relation between \(h\) and \(H\)?
362370
A man throws balls with the same speed vertically upwards one after the other at an interval of \(2 s\). What should be the speed of the throw, so that more than two balls are in the sky at any time? (Take, \(g = 9.8\;m{s^{ - 2}}\) )
1 Any speed less than \(19.6\;m{s^{ - 1}}\)
2 Only with speed \(19.6\;m{s^{ - 1}}\)
3 More than \(19.6\;m{s^{ - 1}}\)
4 Atleast \(9.8\;m{s^{ - 1}}\)
Explanation:
Time taken by ball to reach maximum height, \(v = u - gT\) At maximum height, final speed is zero, i.e. \(v = 0,\,\,\)so,\(u = gT.\) In \(2s,u = 2 \times 9.8 = 19.6\;m{s^{ - 1}}\) If a man throws the ball with velocity of \(19.6\;m{s^{ - 1}},\) then after \(2 s\) it will reach the maximum height. When he throws 2nd ball, 1st is at the top. When he throws third ball, 1 st will come to ground and 2 nd will go at the top. Therefore, only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than \(19.6\;m{s^{ - 1}}\)
362366
A body is projected vertically upwards. The times corresponding to height \(h\) while ascending and while descending are \(t_{1}\) and \(t_{2}\), respectively. Then, the velocity of projection will be (take \(g\) as acceleration due to gravity)
Let \(v\) be initial velocity of vertical projection and \(t\) be the time taken by the body to reach a height \(h\) from ground. Here, \(\quad u = u,a = - g,s = h,t = t\) Using, \(\quad s = ut + \frac{1}{2}a{t^2}\), We have \(h = ut + \frac{1}{2}( - g){t^2}\) or \(g{t^2} - 2ut + 2h = 0\) \(\therefore \quad t = \frac{{2u \pm \sqrt {4{u^2} - 4g \times 2h} }}{{2g}}\) \( = \frac{{u \pm \sqrt {{u^2} - 2gh} }}{g}\) It means \(t\) has two values, i.e. \({t_1} = \frac{{u + \sqrt {{u^2} - 2gh} }}{g}\) \({t_2} = \frac{{u - \sqrt {{u^2} - 2gh} }}{g}\) \({t_1} + {t_2} = \frac{{2u}}{g}\,\,{\rm{or}}\,\,u = \frac{{g\left( {{t_1} + {t_2}} \right)}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362367
A body is dropped from a certain height
1 The time taken to travel first half of the height is less than that for second half
2 The time taken to travel first half of the height is greater than that for second half
3 The time taken to travel first half of the height is equal to that for second half
4 Any one of the above three situations may be correct depending upon the value of height
Explanation:
During first half, velocities are lesser as compared to velocities during second half. So more time is taken in first half.
PHXI03:MOTION IN A STRAIGHT LINE
362368
A small block slides down on a smooth inclined plane, starting from rest at time \(t = 0\). Let \({s_n}\) be the distance travelled by the block in the interval \(t = n - 1\) to \(t = n\). Then , the ratio \(\frac{{{S_n}}}{{{S_{n + 1}}}}\) is
362369
A stone thrown upwards with speed \(u\) attains maximum height \(h\). Another stone thrown upwards from the same point with speed \(2u\) attains maximum height \(H\). What is the relation between \(h\) and \(H\)?
362370
A man throws balls with the same speed vertically upwards one after the other at an interval of \(2 s\). What should be the speed of the throw, so that more than two balls are in the sky at any time? (Take, \(g = 9.8\;m{s^{ - 2}}\) )
1 Any speed less than \(19.6\;m{s^{ - 1}}\)
2 Only with speed \(19.6\;m{s^{ - 1}}\)
3 More than \(19.6\;m{s^{ - 1}}\)
4 Atleast \(9.8\;m{s^{ - 1}}\)
Explanation:
Time taken by ball to reach maximum height, \(v = u - gT\) At maximum height, final speed is zero, i.e. \(v = 0,\,\,\)so,\(u = gT.\) In \(2s,u = 2 \times 9.8 = 19.6\;m{s^{ - 1}}\) If a man throws the ball with velocity of \(19.6\;m{s^{ - 1}},\) then after \(2 s\) it will reach the maximum height. When he throws 2nd ball, 1st is at the top. When he throws third ball, 1 st will come to ground and 2 nd will go at the top. Therefore, only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than \(19.6\;m{s^{ - 1}}\)
362366
A body is projected vertically upwards. The times corresponding to height \(h\) while ascending and while descending are \(t_{1}\) and \(t_{2}\), respectively. Then, the velocity of projection will be (take \(g\) as acceleration due to gravity)
Let \(v\) be initial velocity of vertical projection and \(t\) be the time taken by the body to reach a height \(h\) from ground. Here, \(\quad u = u,a = - g,s = h,t = t\) Using, \(\quad s = ut + \frac{1}{2}a{t^2}\), We have \(h = ut + \frac{1}{2}( - g){t^2}\) or \(g{t^2} - 2ut + 2h = 0\) \(\therefore \quad t = \frac{{2u \pm \sqrt {4{u^2} - 4g \times 2h} }}{{2g}}\) \( = \frac{{u \pm \sqrt {{u^2} - 2gh} }}{g}\) It means \(t\) has two values, i.e. \({t_1} = \frac{{u + \sqrt {{u^2} - 2gh} }}{g}\) \({t_2} = \frac{{u - \sqrt {{u^2} - 2gh} }}{g}\) \({t_1} + {t_2} = \frac{{2u}}{g}\,\,{\rm{or}}\,\,u = \frac{{g\left( {{t_1} + {t_2}} \right)}}{2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362367
A body is dropped from a certain height
1 The time taken to travel first half of the height is less than that for second half
2 The time taken to travel first half of the height is greater than that for second half
3 The time taken to travel first half of the height is equal to that for second half
4 Any one of the above three situations may be correct depending upon the value of height
Explanation:
During first half, velocities are lesser as compared to velocities during second half. So more time is taken in first half.
PHXI03:MOTION IN A STRAIGHT LINE
362368
A small block slides down on a smooth inclined plane, starting from rest at time \(t = 0\). Let \({s_n}\) be the distance travelled by the block in the interval \(t = n - 1\) to \(t = n\). Then , the ratio \(\frac{{{S_n}}}{{{S_{n + 1}}}}\) is
362369
A stone thrown upwards with speed \(u\) attains maximum height \(h\). Another stone thrown upwards from the same point with speed \(2u\) attains maximum height \(H\). What is the relation between \(h\) and \(H\)?
362370
A man throws balls with the same speed vertically upwards one after the other at an interval of \(2 s\). What should be the speed of the throw, so that more than two balls are in the sky at any time? (Take, \(g = 9.8\;m{s^{ - 2}}\) )
1 Any speed less than \(19.6\;m{s^{ - 1}}\)
2 Only with speed \(19.6\;m{s^{ - 1}}\)
3 More than \(19.6\;m{s^{ - 1}}\)
4 Atleast \(9.8\;m{s^{ - 1}}\)
Explanation:
Time taken by ball to reach maximum height, \(v = u - gT\) At maximum height, final speed is zero, i.e. \(v = 0,\,\,\)so,\(u = gT.\) In \(2s,u = 2 \times 9.8 = 19.6\;m{s^{ - 1}}\) If a man throws the ball with velocity of \(19.6\;m{s^{ - 1}},\) then after \(2 s\) it will reach the maximum height. When he throws 2nd ball, 1st is at the top. When he throws third ball, 1 st will come to ground and 2 nd will go at the top. Therefore, only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than \(19.6\;m{s^{ - 1}}\)
