A ball is thrown vertically upwards, with a speed of 10m/s from the top of a tower 200m high and another is thrown vertically downwards, with the same speed simultaneously. What is the time difference between them in reaching ground?
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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
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
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