Question

A ball is thrown straight up with an initial velocity of 6.4 m/s. It travels for 0.64 seconds, and has a change of position of 2.05 meters. What is the ball’s final velocity?

Answer 1

V = 0.9 m/s

Step-by-step explanation:

The parameters given are:

Initial velocity U = 6.4 m/s

Time t = 0.64s

Height h = 2.05 m

To find the final velocity, let us use third equation of motion

V^2 = U^2 - 2gH

Since the ball is going upward, g will be negative

Substitute all the parameters into the formula

V^2 = 6.4^2 - 2 × 9.8 × 2.05

V^2 = 40.96 - 40.18

V^2 = 0.78

V = sqrt( 0.78)

V = 0.883 m/ s

V = 0.9 m/ s approximately