A ball is thrown horizontally from a height of 18.18 m and hits the ground with a speed that is 4.0 times its initial speed. What was the initial speed?

Answer 1

The vertical velocity component when it hits the ground will be

Vy = sqrt(2gH)= 17.89 m/s

Speed at that time = sqrt(Vy^2 + Vx)^2

The initial speed was Vx. That component of the velocity remained unchanged.

sqrt(Vy^2 + Vx)^2 = 4 Vx
Vy^2 = 15 Vx^2
Vx = Vy/sqrt15 = 4.619 m/s
= initial speed

Final speed = 18.48 m/s

Answer 2

A ball with an initial speed of = 18.5 m/s collides elastically with two identical balls whose centers are on a line perpendicular to the initial velocity and that are initially in contact with each other. The first ball is aimed directly at the contact point and all motion is frictionless.



What is the speed of ball 1 after the collision?

What is the speed of ball 2 after the collision?