Acomputer-controlled ball launcher on a ledge at height h = 8 m above level ground ejects a ball at initial speed v_sub_0 = 8.2 m/s at an angle of θ_sub_1 = 35 degrees above the horizontal. the computer then commands the launcher to change its launch angle to a specific value θ_sub_2 below the horizontal, and to launch a second ball exactly 1.00 seconds after the first ball at a different speed from the first one. assume the balls are launched from the same height, move in the same plane, and that air resistance can be ignored. the launch of the second ball is arranged such that the second ball collides with the first ball at a height h/2 above the level ground. this requires the second ball to have a specific initial speed and launch angle. what is the required initial speed?

21 m/s

Explanation:

For the first ball, in the x direction:

x = x₀ + v₀ t + ½ at²

x = 0 + (8.2 cos 35) t + ½ (0) t²

x = 6.72t

In the y direction:

y = y₀ + v₀ t + ½ at²

y = 8 + (8.2 sin 35) t + ½ (-9.8) t²

y = 8 + 4.70t − 4.9t²

When y = 4:

4 = 8 + 4.70t − 4.9t²

4.9t² − 4.70t − 4 = 0

Solve for t with quadratic formula:

t = [ 4.70 ± √((-4.70)² − 4(4.9)(-4)) ] / 9.8

t = (4.70 ± 10.0) / 9.8

t = 1.50

Therefore:

x = 6.72t

x = 10.1

Now, for the second ball in the x direction:

x = x₀ + v₀ t + ½ at²

x = 0 + (v₀ cos (-θ)) (t − 1) + ½ (0) (t − 1)²

x = v₀ cos θ (t − 1)

And in the y direction:

y = y₀ + v₀ t + ½ at²

y = 8 + (v₀ sin (-θ)) (t − 1) + ½ (-9.8) (t − 1)²

y = 8 − v₀ sin θ (t − 1) − 4.9(t − 1)²

When t = 1.50, x = 10.1 and y = 4:

10.1 = v₀ cos θ (1.50 − 1)

v₀ cos θ = 20.1

4 = 8 − v₀ sin θ (1.50 − 1) − 4.9(1.50 − 1)²

4 = 6.76 − 0.50 v₀ sin θ

v₀ sin θ = 5.49

Using Pythagorean theorem:

v₀² = (v₀ cos θ)² + (v₀ sin θ)²

v₀² = (20.1)² + (5.49)²

v₀ = 20.8

Rounded to two significant figures, the required initial speed is 21 m/s.

measuring things 1 litre of water