Question

21. A stone is dropped from top of tower of height
120 m and at the same instant another stone is
thrown upward with a velocity of 40 m/s from
bottom of tower. The time after which the two
stone will meet is​

Answer 1

Hey Dear,

◆ Answer -

t = 3 s

● Explanation -

Let t be the time after which two stones will meet each other.

At time t, distance covered by stone dropped from above will be -

s1 = ut + 1/2 at^2

s1 = 0×t + 1/2 × 10 × t^2

s1 = 5t^2

At time t, distance covered by stone thrown from below will be -

s2 = ut + 1/2 at^2

s2 = 40×t + 1/2 × (-10)t^2

s2 = 40t - 5t^2

Total distance covered by two stones will be height of tower.

s1 + s2 = h

5t^2 + 40t - 5t^2 = 120

40t = 120

t = 3 s

Therefore, two stones will meet each other after 3 s.

Thanks dear...