Question

A blueberry pie is shot out of a pie cannon from a height of 6m and with a velocity of 60 m/s. How many seconds will pass before the pie hits the ground?

Answer 1

Answer: 1.106 s

Step-by-step explanation:

This situation is related to projectile motion and one the equation that models the height of the blueberry pie in time is:

$y=y_{o}+V_{o}sin(\theta) t-\frac{1}{2}gt^{2}$

Where:

$y=0 m$ is the blueberry pie final height (when it hits the ground)

$y_{o}=6 m$ is the blueberry pie initial height

$V_{o}=60 m/s$ is the blueberry pie initial velocity

$\theta=0 \°$ is the angle, assuming the pie was shot horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

Rewriting the equation:

$0=y_{o}-\frac{1}{2}gt^{2}$

Isolating $t$:

$t=\sqrt{\frac{2y_{o}}{g}}$

$t=\sqrt{\frac{2(6 m)}{9.8 m/s^{2}}}$

Finally:

$t=1.106 s$