Question

a body released from top of the tower of height H meters it takes time t to reach the ground . where is the body t÷2 time after the release:

Answer 1

you know the formula...

H= uT + 0.5g(T^2)

So, the body will take time t, which is equal to

$\sqrt{(2h \div g)}$

Now 'H' is directly proportional to 'T^2' if u=0 and g is constant which is the situation here.

So, after (t÷2)s it will cover (H÷4) meters.

$\frac{h2}{h1} = ( \frac{t2}{t1}) {}^{2}$

now.. (t2÷t1)^2 = ((t÷2)÷ t)^2
=(1÷4)
therefore, h2= h1÷4

so, the body will remain at the height (3H÷4) from the ground as it crosses distance (H÷4) from the top.

Hope you're helped by the answer.