Question

a stone is allowed to fall from the top of a tower 40m high and at the same time a stone is thrown upwards with a velocity of 10m/s . find when and where the two stones meet
( pls explain briefly )

Answer 1

Answer:

80 m

Explanation:

Height covered by the falling stone = S₁

∴ S₁ = ut + (1/2) gt²

      = 0 * t + (1/2) * (10) t²

     = 5t²

Distance covered by the stone thrown upward = S₂

g = -10 m/s

u = 10 m/s.

S₂ = ut + (1/2) gt²

   = 10 * t + (1/2) * (-10) * t²

  = 10t - 5t²

Given,

Total height = 40 m

∴ S₁ + S₂ = 40 m

⇒ 5t² + (10t - 5t²) = 40

⇒ 10t = 40

⇒ t = 4 seconds

Place t = 4 in any equation, we get

S₁ = 5t²

  = 5(4)²

 = 80 m

Therefore,

The two stones will meet after 4 seconds when the falling stone has covered a distance of 80 m.

Hope it helps!