Problem 3-21
A ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0 m from the
base. What was the ball’s initial speed?
Since v0 is horizontal, v0x = v0 and v0y = 0.
Equations for Uniform Acceleration
x = x0 + v0xt
0 = y0 + ½ ay t2
x = v0t
𝑣0 =
y = y0 + v0yt + ½ ayt2
𝑥
𝑡
24.0
𝑣0 = 3.0304
We need time to
calculate v0, so use
other equation first.
−2𝑦0
𝑡=√
𝑎𝑦
−2(45.0)
𝑡=√
−9.8
t = 3.0304 s
v0 = 7.9195 m/s
v0 = 7.92 m/s
The ball’s initial speed was 7.92 m/s.
Plug this
value in for t
in other
equation.