Question

A baseball is hit with a speed of 27.0 m/s at an angle of 45.0o. It lands on the flat roof of a 13.0 m tall nearby building. If the ball was hit when it was 1.0 m above the ground, what horizontal distance does it travel before it lands on the building?

Answer 1

Answer: 10.175 m

Explanation:

This situation is related to parabolic movement with constant acceleration (due gravity) and can be modeled by the following equations:

$V^{2}=V_{o}}^{2}-2gy$    (1)

$V=V_{o}-g.t$    (2)

$x=V_{o}cos\theta t$    (3)

Where:

$y=13m$  is the baseball's final height (when it lands on the building's roof)

$V_{o}=27\frac{m}{s}$ is the  baseball's initial velocity

$V$ is the baseball's final velocity

$t$ is the time the parabolic movement lasts  until the baseball lands on the roof

$g=9.8 \frac{m}{s^{2}}$  is the acceleration due to gravity

$x$ is the horizontal distance  the baseball travels

$\theta=45\°$ is the angle at which the baseball was hit

Let's begin finding $V$  with (1):

$V=\sqrt{V_{o}}^{2}-2gy}$    (4)

$V=\sqrt{(27\frac{m}{s})^{2}-2(9.8 \frac{m}{s^{2}})(13 m)}$    (5)

$V=21.776 \frac{m}{s}$    (6)

Substituting (6) in (2):

$21.776 \frac{m}{s}=27\frac{m}{s}-(9.8 \frac{m}{s^{2}}).t$    (7)

Isolating $t$:

$t=0.533 s$    (8)

Substituting (8) in (3):

$x=27\frac{m}{s}cos(45\°) 0.533 s$    (9)

Finally:

$x=10.175 m$