a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v

Question

3 years ago

13

erlocity of 50m/s by a person standing at the foot of the building.find the time after which the two stones meet.

1 answer:

solniwko [45] · Answer 1 · 2022-05-19T00:02:23+03:00

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

The height of the building, h = 200 m

The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

S = ut + ½ gt²

Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)

Adding these two equations,

x + (S - x) = U tₓ

S = U tₓ

200 = 50 tₓ

∴ tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s