Question

when we throw a ball with a speed of 40 metre per second after 2 second we through the another ball in which height both the ball will collide?​

Answer 1

Explanation:

X be the distance from the ground where the collision takes place then

Distance from top of tower up to the collision point =40-X

let collision took place after time t then

X=20t-1/2×g×t^2……(1)(upward motion)

40-X=20t+1/2g×t^2….(2)(downward motion)

Adding (1) and (2) we get

40=40t

Which gives t=1 second

Putting value of t in (1) we get

( taking :g= 9.8 m /s^2)

X=15.1 metre

So collision takes at 15.1 metres from ground or 40–15.1=24.9 metres from top.

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Answer 2

Answer:

what is the speed of second ball? And from where (up/down) the ball was thrown?