Question

A ball is thrown straight up with an initial speed of 60 m/s. How high does it go? Assume the accelera-
tion of gravity is 10 m/s22


2)How long is it in the air?

Answer 1

1. Vf^2 + 2ad = 0,
(60)^2 + 2 * (-10)d = 0,
3600 - 20d = 0,
-20d = -3600,
d = 180 m upward.

2. d = Vo*t + 0.5at^2 = 180 m upward,
60t + 0.5*(-10)t^2 = 180,
60t - 5t^2 = 180,
Rearrange terms:
-5t^2 + 60t = 180,
Divide both sides by -5:
t^2 - 12t = - 36,
t^2 - 12t + 36 = 0,
Factor and get:
(t - 6) (t - 6) = 0,
t - 6 = 0,
t = 6 s = t(up).

d = 0.5at^2 = 180 m downward,
0.5 * 10t^2 = 180,
5t^2 = 180,
t^2 = 36,
t = 6 s downward.

t(up) + t(down) = 6 + 6 = 12 s in air.