Question

A baseball is thrown into the air and follows a parabolic path given by the equation s = -16t 2 + v0t, where s is feet above ground, t is the time in seconds and v0 is the initial velocity. If the ball is thrown with an initial velocity of 64 feet per second, how high will it travel?
16 ft.
64 ft.
128 ft.

Answer 1

Answer: 64 ft

Step-by-step explanation:

We are given the following equation that models the baseball's parabolic path:

$s=16t^{2}+V_{o}t$ (1)

Where:

$s$ is the ball maximum height

$t$ is the time

$V_{o}=64 ft/s$ is the initial height

With this information the equation is rewritten as:

$s=16t^{2}+(64)t$ (2)

Now, we have to find $t$, and this will be posible with the following formula:

$V=V_{o}-gt$ (3)

Where:

$V=0 ft/s$ is the final velocity of the ball at the point where its height is the maximum

$g=32 ft/s^{2}$ is the acceleration due gravity

Isolating $t$:

$t=\frac{V_{o}}{g}$ (4)

$t=\frac{64 ft/s}{32 ft/s^{2}}$ (5)

$t=2 s$ (6)

Substituting (6) in (2):

$s=16 ft/s^{2}(2 s)^{2}+(64 ft/s)(2 s)$ (7)

Finally:

$s=64 ft$