Question

A ball is launched straight up from the edge of a cliff that is 250 m high. The ball's motion is given by s = 35t - 4.9t^2 where s is the ball's height in meters above the cliff top at time t seconds.

With what velocity does the ball strike the ground at the base of the cliff?
Answer: -285 m/sec

Is that correct?
Nope. That's way too fast.

How long does it take to fall to a height of -250m?

35t - 4.9t^2 = -250
t = 11.56

Now,

v = 35 - 9.8t = -78.26

Related Questions

Kevin stands at the edge of a cliff, holding one ball in each hand. He throws one of the balls strai... A projectile is launched straight up at 55.5 m/s from a height of 73.5 m, at the edge of a sheer cli... You stand at the edge of a cliff, holding a ball out over the edge. You give the ball an initial vel... A ball is thrown straight up at 13.0 m/s near the edge of a 30.0-meter high cliff. The ball lands at...