Question

A ball is projected from the bottom of a tower and is found to go above the tower and is caught by the thrower at the bottom of the tower after a time interval $t_1$. An observer at the top of the tower sees the same ball go up above him and then come back at this level in a time interval $t_2$. The height of the tower is

• A. $\frac{g(t_1-t_2)^2}{8}$
• B. $\frac{1}{2}gt_1t_2$
• C. $\frac{g}{8}(t_1^2-t_2^2)$
• D. $\frac{gt_1t_2}{8}$

Answer 1

Answer: (C)

$\frac{g}{8}(t_1^2-t_2^2)$

Answer 2

Let $u_y$ be the initial vertical velocity. The total time of flight $t_1$ for the ball to return to the ground is given by $t_1 = \frac{2u_y}{g}$, so $u_y = \frac{gt_1}{2}$.

Let $h$ be the height of the tower. The time interval $t_2$ is the time taken for the ball to go up from height $h$ and come back to height $h$. This is a projectile motion segment where the initial and final heights are the same. If $v_{up}$ is the velocity at height $h$ going upwards, then $t_2 = \frac{2v_{up}}{g}$, so $v_{up} = \frac{gt_2}{2}$.

The time taken to reach height $h$ from the ground is $t_{up}$. Using the kinematic equation $v = u - gt$, we have $v_{up} = u_y - gt_{up}$. Substituting the expressions for $v_{up}$ and $u_y$: $\frac{gt_2}{2} = \frac{gt_1}{2} - gt_{up}$ Dividing by $g$: $\frac{t_2}{2} = \frac{t_1}{2} - t_{up}$ $t_{up} = \frac{t_1 - t_2}{2}$

Now, we can find the height $h$ using the equation $h = u_y t_{up} - \frac{1}{2}gt_{up}^2$: $h = \left(\frac{gt_1}{2}\right) \left(\frac{t_1 - t_2}{2}\right) - \frac{1}{2}g \left(\frac{t_1 - t_2}{2}\right)^2$ $h = \frac{gt_1(t_1 - t_2)}{4} - \frac{g(t_1 - t_2)^2}{8}$ To combine these terms, we find a common denominator: $h = \frac{2gt_1(t_1 - t_2) - g(t_1 - t_2)^2}{8}$ Factor out $g(t_1 - t_2)$: $h = \frac{g(t_1 - t_2)[2t_1 - (t_1 - t_2)]}{8}$ $h = \frac{g(t_1 - t_2)(t_1 + t_2)}{8}$ Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$: $h = \frac{g(t_1^2 - t_2^2)}{8}$