Question

A ball is released from the top of tower of height h meter .it takes T second to reach the ground.what is the position of the ball in T/2seconds

Answer 1

Topic :- Motion under gravity

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

A ball is released from the top of tower of height h meter. That means Initial Velocity (u) will be 0 in Time (t) = 0 and Time taken to reach ground (T) will be $\sf \sqrt{\dfrac{2H}{g}}$. We need to find the position of the ball in T/2 seconds.

• $\bf Let\: t_1 = \dfrac{T}{2}\: s$

By using second kinematical equation of motion we have :-

$\longrightarrow \sf H_1 = u + \dfrac{1}{2} gt_1^2$

$\longrightarrow \sf H_1 = u + \dfrac{1}{2} g \bigg( \dfrac{T}{2} \bigg)^2$

$\longrightarrow \sf H_1 = 0 + \dfrac{1}{2} g \bigg( \dfrac{T}{2} \bigg)^2$

$\longrightarrow \sf H_1 = \dfrac{1}{2} g \bigg( \dfrac{T}{2} \bigg)^2$

$\longrightarrow \sf H_1 = \dfrac{1}{2} g \dfrac{T^{2} }{4} \qquad$

$\longrightarrow \sf H_1 = \dfrac{1}{2} \times g \times \dfrac{1 }{4} \times T^{2} \qquad ...(1)$

But we know that time taken to reach ground (T) will be

$\bullet \: \sf T = \sqrt{\dfrac{2H}{g}}$

$\bullet \: \sf T^{2} = \dfrac{2H}{g} \qquad...(2)$

Substituting the value of T from equation (2) to equation (1) :

$\longrightarrow \sf H_1 = \dfrac{1}{2} \times g \times \dfrac{1 }{4} \times \dfrac{2H}{g}$

$\longrightarrow \sf H_1 = \dfrac{1}{2} \times g \times \dfrac{2H }{4g}$

$\longrightarrow \underline{ \boxed{\sf H_1 = \dfrac{H }{4}}}$

Also,

$\longrightarrow \sf H = H_1 + H_2$

$\longrightarrow \sf H_2 = H - H_1$

$\longrightarrow \sf H_2 = H - \frac{H}{4}$

$\longrightarrow \sf H_2 = \frac{ 4 H - H}{4}$

$\longrightarrow \underline{ \boxed{ \sf H_2 = \frac{ 3H}{4}}}$