Question

A ball is thrown straight up into the air with a velocity of 12 m/s. Draw a motion diagram for the ball and then give as much quantitative information as you can about the motion

Answer 1

Answer:

find the diagram in the attachment.

Explanation:

Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.

considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:

(vf)^2 = (vi)^2 + 2×g×Δy

vf = 0 m/s, at the highest point in the upward motion, then:

0 = (vi)^2 + 2×g×Δy

-(vi)^2  = 2×g×Δy

Δy = [-(vi)^2]/2×g

Δy = [-(-12)^2]/(2×9.8)

Δy = - 7.35 m

then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown