# While on planet Soccerball, an object is thrown vertically upward with an initial velocity of 5.0 m/s. If this object returns to the point of release in 3.0 s, what is the acceleration of a free falling object on this planet?

Apr 20, 2018

$- 3.33 \frac{m}{s} ^ 2$

#### Explanation:

So for this one you have to first find the time it takes for the ball to reach the peak of its parabola. In all cases of an equal height projectile problem, where the projectile starts and finishes at the same elevation, the time it takes to complete its trajectory is always twice the time it takes to reach the top of its parabola.

Since it takes 3.0 seconds for the projectile to rise and fall, it takes only 1.5 seconds for the projectile to rise.

You can think of this time as the time required for the acceleration due to gravity of the planet (which works against your upwards velocity) to take away all your upwards velocity.

Note that at the top of your parabola, the upwards velocity will equal zero.

Now you have both a $\Delta V$ and a $\Delta T$ so you can solve for $a$ using the following equation.

$a = \frac{\Delta V}{\Delta T}$

$a = \frac{{V}_{f} - {V}_{i}}{1.5}$

$a = \frac{0 - 5}{1.5}$

$a = - 3.33$

Your acceleration due to gravity on Planet Soccer Ball is $- 3.33 \frac{m}{s} ^ 2$

This problem assumes no air resistance.