Question

Two boys are throwing a baseball back and forth. The ball is 4 ft above the ground when it leaves one child’s hand with an upward velocity of 36 ft/s. If acceleration due to gravity is –16 ft/s2, how high above the ground is the ball 2 s after it is thrown?

Answer 1

@Shadow

Answer 2

Hello @Camila

Here you will need to be using a kinematic equation.

\[d_{y} = v_{i} t + \frac{ 1 }{ 2 }at^2\]

This basically says: the distance of an object in the y direction is equal to the objects initial velocity multiplied by time plus half of the objects acceleration multiplied by the square of time.

Answer 3

good job shad

Answer 4

Since there is an initial height of 4 ft, you will need to account for that in the right hand side of the equation, so simply add 4. If you need help identifying where all the numbers go, just ask in chat as I have to go.

Answer 5

what he said