Question

From the top of the tower a stone is dropped if it covers 24.5 in the last second of it motion height of the tower is

Answer 1

Explanation:

By newtons law

s = vt + 1/2 at^2.

24.5 = v(1)+ 1/2g(1)^2

24.5–1/2g=v

Taking g = 9.8m/s^2

v=19.6 m/s

Now v^2 - u^2 =2(9.8)S

19.6×19.6 = 2(9.8)S

As the stone was dropped so u=0

S=19.6

So total height from where the stone was dropped =S+s =19.6+24.5 =44.1 m

Since the stone travels 24.5 m in the last second of its journey

S=ut +1/2at^2

24.5 = u(1) + 1/2(9.8)(1*1)

24.5–4.9 = u

u = 19.6 m/s

V^2 -u^2= 2as

19.6^2 - 0^2 = 2(9.8)s

S= 19.6 m

Total height is 19.6+ 24.5 = 44.1m