Question

a stone is dropped from the top of the building 200m high and at the same time another a stone is projected vertically upward from the ground with a velocity 50 metre per second find where and when to will meet​

Answer 1

Answer:

Let the two stones meet at a distance of x mfrom the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.

For the stone that is dropped its initial velocity u=0ms

−1

; displacement s = x and acceleration = acceleration due to gravity (g).

Using s=ut+

2

1

at

2

,

wegetx=(0)t+

2

1

gt

2

→(1).

For the stone that is projected vertically upwards, its initial velocity, u=−50ms

−1

; displacement s =-(200-x) and acceleration a =g.

using s=ut+

2

1

at

2

,

weget−(200−x)=−50×t+1/2gt

2

200=50t−1/2gt

2

+x→(1)(2)

Fromtheequations(1)and(2)

wehave200=50t−1/2gt

2

+1/2gt

2

⇒200=50∴t=4s