Physics, asked by adhithyavemula, 1 year ago

A ball is thrown vertically upwards with a speed of 10 metre per second from the top of a tower 200 height and other is thrown vertically downwards with the same speed simultaneously. the time difference between them and reaching ground in seconds if gravity is taken as 10 metre per second square

Answers

Answered by kaalipavan
2

Answer:

Explanation:

The equation for an object’s height is h(t)=g/2 t² + vt + c, where h(t)=objects height at t seconds, g is gravity, v is initial velocity, and c is initial height. So, for the first ball, it hits the ground when h(t)=0. So:

0=g/2 t²+10 t +200

For the second ball, we have:

0=g/2 t²-10t+200

Solving for t in both equations, we get:

0=-4.9t²+10t+200

4.9t²-10t-200=0

t=7.5

and

0=-4.9t²-10t+200

4.9t²+10t-200=0

t=5.5

The difference is 7.5–5.5=2 secs

Similar questions