(B) Let the initial velocity be $u$. The total time of flight is $T = t_1 + t_2$. Since the ball returns to the ground,the total displacement is $0$.

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(B) Let the initial velocity be $u$. The total time of flight is $T = t_1 + t_2$. Since the ball returns to the ground,the total displacement is $0$. Using the equation of motion $S = ut + \frac{1}{2}at^2$ with $a = -g$:$0 = u(t_1 + t_2) - \frac{1}{2}g(t_1 + t_2)^2$$u = \frac{1}{2}g(t_1 + t_2)$Now,the height $h$ reached at time $t_1$ is given by:$h = ut_1 - \frac{1}{2}gt_1^2$Substituting the value of $u$:$h = \left[\frac{1}{2}g(t_1 + t_2)\right]t_1 - \frac{1}{2}gt_1^2$$h = \frac{1}{2}gt_1^2 + \frac{1}{2}gt_1t_2 - \frac{1}{2}gt_1^2$$h = \frac{1}{2}gt_1t_2$

Class 11 Physics Motion in Straight Line Motion Under Gravity

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$A$ ball is projected vertically upwards from the ground. It reaches a height $h$ in time $t_1$,continues its motion,and then takes a time $t_2$ to reach the ground. The height $h$ in terms of $g, t_1$,and $t_2$ is ($g =$ acceleration due to gravity).

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A
$\frac{1}{2} \frac{gt_1}{t_2}$
B
$\frac{1}{2} gt_1 t_2$
C
$g t_1 t_2$
D
$2 gt_1 t_2$

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Motion in Straight Line

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$A$ stone is dropped from a height $h$. Simultaneously,another stone is thrown up from the ground which reaches a height $4h$. The two stones cross each other after time:

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$A$ ball is dropped downwards. After $1$ second,another ball is dropped downwards from the same point. What is the distance between them after $3$ seconds?

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$A$ ball of mass $m$ is thrown vertically upwards. Assume the force of air resistance has magnitude proportional to the velocity,and direction opposite to the velocity. At the highest point,the ball's acceleration is

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$A$ man throws a ball vertically upward and it rises through $20 \; m$ and returns to his hands. What was the initial velocity $(u)$ of the ball and for how much time $(T)$ it remained in the air? $[g = 10 \; m/s^2]$

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$Assertion$ : Two balls of different masses are thrown vertically upward with the same speed. They will pass through their point of projection in the downward direction with the same speed.
$Reason$ : The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball.

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$A$ ball is projected vertically upwards from the ground. It reaches a height $h$ in time $t_1$,continues its motion,and then takes a time $t_2$ to reach the ground. The height $h$ in terms of $g, t_1$,and $t_2$ is ($g =$ acceleration due to gravity).

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$A$ stone is dropped from a height $h$. Simultaneously,another stone is thrown up from the ground which reaches a height $4h$. The two stones cross each other after time:

$A$ ball is dropped downwards. After $1$ second,another ball is dropped downwards from the same point. What is the distance between them after $3$ seconds?

$A$ ball of mass $m$ is thrown vertically upwards. Assume the force of air resistance has magnitude proportional to the velocity,and direction opposite to the velocity. At the highest point,the ball's acceleration is

$A$ man throws a ball vertically upward and it rises through $20 \; m$ and returns to his hands. What was the initial velocity $(u)$ of the ball and for how much time $(T)$ it remained in the air? $[g = 10 \; m/s^2]$

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