A baseball is thrown into the air with an initial velocity of 64 ft/sec. How high will it travel and how long will it take to reach its maximum height?
-
I guess it's thrown straight up. This is a constant acceleration problem with a = g = -32 ft./s/s and v0 = 64 ft./s.
It's easier to solve the first part if you solve the second part first. When it reaches its maximum height, recognize that v = 0.
v = v0 + at
0 = 64 + -32t
32t = 64
t = 2s
Now use
y = v0t + (1/2)at^2
y = 64 * 2 + (1/2) * -32 * 2^2
y = 128 + (1/2) * -32 * 4
y = 128 - 64
y = 64 ft.
It's easier to solve the first part if you solve the second part first. When it reaches its maximum height, recognize that v = 0.
v = v0 + at
0 = 64 + -32t
32t = 64
t = 2s
Now use
y = v0t + (1/2)at^2
y = 64 * 2 + (1/2) * -32 * 2^2
y = 128 + (1/2) * -32 * 4
y = 128 - 64
y = 64 ft.