From the top of a building 160 m high, a ball is
dropped and at the same time a stone is projected
vertically upward from the ground with a velocity of
40 ms-1. Find when and where the stone meet.
Take g = 10 m/s.
Answers
Answered by
2
Answer:
Let us consider the distance they meet at to be 'x' metres from ground.
Initial velocity of ball=0 m/s
Initial velocity of stone = 40m/s
The second law of motion gives us
(160-x)+x= (0+ g*t^2 /2) + ( 40t-g*t^2 /2)
160= 40t
t=4 s
x= 40t - g*t^2 /2
= 160 - 9.8*16/2
=81.6 m
They meet at 81.6 metres from the ground, 4 seconds after their initial launch.
Similar questions