A laboratory projectile launcher ontop of the edge of a table fires a metal ball horizontally 1.50m above the floor and hits the floor 2.50m?

Assuming projectile is released from launcher at the top edge of table and the height of the projectile from the floor at this release point = h = 1.50 m and the projectile hits the floor at a horizontal distance from the edge of table = d = 2.50 m then...

time for projectile to Free-Fall to the floor = t = √(2h/g) = √[2(1.50)/9.81] ≈ 0.533 s ANS a.

d = Vh(t) {where Vh = initial = constant horizontal speed of projectile}

Vh = d/t = 2.50/0.53300 ≈ 4.69 m/s ANS B.

Since the ball is launched horizontally, its initial vertical velocity is 0 m/s. Let’s use the following equation to determine the time for the ball to fall 1.50 meters.

d = vi * t + ½ * a * t^2, vi = 0

1.50 = ½ * 9.8 * t^2

t = √(1.50 ÷ 4.9)

This is approximately 0.55 second. During this time, the ball moves 2.50 meters horizontally. Let’s use the following equation to determine the ball’s initial velocity

d = v * t

2.50 = v * √(1.50 ÷ 4.9)

v = 2.50 ÷ √(1.50 ÷ 4.9)

This is approximately 4.52 m/s.