Question

Q.10
A ball is dropped from the top of 100 m high
tower on a planet. In last / s before hitting the
ground. It cover a distance of 19 m.
Acceleration due to gravity (m/s)near
surface on the planet is :
(1) 8 m/s (2) 28 m/s
(3) 10 m/s? (4) none of these​

Answer 1

Acceleration due to gravity (m/s) near surface on the planet is:

(1) 8 m/s²

Initially,

u=0, x=100, a=g′, t=t₀

Now, x = u t + (1/2)at²

Substituting the values, we get

100 = 0×t + (1/2) ​×g′ t²₀

g′ t²₀ = 200                                           (eq1)

Now, before hitting the ground,

u=0, x= (100−19)= 81, a=g′, t= (t₀−(1/2))

x=u t+(1/2)​at²

Substituting the values we get,

81 = 0×t + (1/2)×g′×(t₀− (1/2)​)²

81 = (g′/2) ​(t₀−(1/2))²

g′(t₀−(1/2))²=162                                     (eq2)

from (eq1) and (eq2)

g′t²₀​​ / g′(t₀​−(1/2)) = 200/162​

                          = 100/81​

Now, t₀/​ t₀-(1/2) ​​= 10/9​

9t₀ = 10t₀−5

∴ g′ =200​/ t²₀

     ​​  =200/5²

       ​=8m/s²