Question

You throw a ball straight up into the air at a velocity of 15 m/s. You want to know how high above your hand the ball will be at exactly 2.5 sec after you released it. You must SHOW ALL WORK to receive credit!!! Of the 5 motion variables - Vi, Vf, a, ΔX, and T, identify which one you are solving for and the magnitudes and directions of all of the others that are known: Which of the motion equations is best to use to determine the height of the ball? How high is the ball above your hand at 2.5 sec after you throw it?

Answer 1

6.875m

Step-by-step explanation:

Given

Velocity = 15m/s

Time = 2.5secs

Required

Distance above your hand the ball will be at exactly 2.5secs

using the equation of motion:

ΔX = ViT + 1/2aT²

ΔX is the distance

Vi is the velocity of the ball in air

a is the acceleration due to gravity

T is the time taken

Since the ball is thrown straight up, a = -g

ΔX = ViT - 1/2gT²

Substitute

ΔX = 15(2.5) - 1/2(9.8)(2.5)²

ΔX = 37.5-30.625

ΔX = 6.875m

Hence the ball will be 6.875m above your hand at 2.5secs after you throw it.