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?

1 answer

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