Question

a body is released from the top of a tower of height H metre it takes t seconds to reach the ground where is the ball at the time T by 2 seconds​

Answer 1

$\huge\pink{Solution}$

Height = h

u = 0 m/s

Time = t

We know that,

$s = ut + \frac{1}{2}g {t}^{2}$

So,

$h = 0 \times t + \frac{1}{2} \times g \times {t}^{2}$

$h1 = \frac{1}{2} g {t}^{2}$

$\large\purple{Now}$

$time = \frac{t}{2}$

So,

$h2 = \frac{1}{2} \times g \times { \frac{t}{2} }^{2}$

$h2 = \frac{1}{2} \times g \times { \frac{t}{4} }^{2}$

$\frac{h1}{h2} = \frac{4}{1}$

$h2 = \frac{h1}{4}$

Hence, the ball will be at the height of

$h - \frac{h}{4} = \frac{3}{4} h$

$\large\red{Hope}$$\large\red{it}$$\large\red{helps}$