Question

A stone is dropped from the top of a tower of height h=125m. After 1 second another stone is dropped from the balcony 'x' m below the top of the tower
If both reach the ground simultaneously then the value of 'x' is (g = 10ms 2)
(A) 80 m
(B) 45 m
(C) 120 m
(D) 5 m​

Answer 1

Answer:

45 m

Explanation:

hope you got it and it helps you.

Answer 2

Answer:

(B) x = 45 m

Explanation:

Height of tower = 125 m (s₁)

Let Stone 1 = St₁ and Stone 2 = St₂

Initial velocity of St₁ = 0 m/s

Acceleration due to gravity = 10 m/s²

Initial velocity of St₂ = 0 m/s

Distance which St₂ has to cover = 125-x (s₂)

By 2nd equation of motion,

s₁ = u t + 1/2 at²

125 = (0) (t) + 1/2 × 10 × t²

125 = 5t²

25 = t²

5 s = t

∴ Stone 1 takes 5 seconds to cover 125 m. Now, if Stone 2 has to reach simultaneously with Stone 1, then it must travel the required distance (125-x) in 5-1 = 4 seconds.

s₂ = u t + 1/2 at²

125-x = (0) (4) + 1/2 × 10 × 4²

125-x = 5 × 16

-x = 80-125

x = 45 m                                        (ANS.)

THEREFORE, THE SECOND STONE WAS RELEASED FROM A HEIGHT OF 45 M BELOW THE HIGHEST POINT OF THE TOWER.