A ball is thrown straight up. If the launch velocity is 15 m/s, at what SPEED will the ball return to the thrower’s hand? a) -15 m/s
b) -9.8 m/s
c) 15 m/s
d) 0 m/s

Vf = 15 m/s

Explanation:

First we consider the upward motion of ball to find the height reached by the ball. Using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = -9.8 m/s² (negative sign for upward motion)

h = height =?

Vf = Final Velocity = 0 m/s (Since, ball momentarily stops at highest point)

Vi = Initial Velocity = 15 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (15 m/s)²

h = (-225 m²/s²)/(-19.6 m/s²)

h = 11.47 m

Now, we consider downward motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 11.47 m

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

2(9.8 m/s²)(11.47 m) = Vf² - (0 m/s)²

Vf = √(224.812 m²/s²)

Vf = 15 m/s

friction

explanation:

they us to move forward and backwards