Question

1) A 85.0 kg football player leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 21.2 inches off the ground. He hits the ground 1.11 inches away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball travelling?

2)You would like to shoot an orange in a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. On your first try, you fire the arrow at 36.0 m/s at an angle of 30.0° above the horizontal from a height of 1.40 m while standing 52.0 m away. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has traveled the 52.0 m horizontally to the tree?

Answer 1

here

The time needed to free fall a height of 21.2 inch=1.767 ft is
s = a * t^2 / 2

g = 9.8m/s^2 = 32.15 ft/s^2

1.767 = 32.15 * t^2 / 2

t = 0.33 s

In that time, he makes a horizontal displacement of 1.11 in (=0.0925 ft).
So his horizontal velocity was

r = v * t

0.0925 = v * 0.33

v = 0.28 ft/s

He must have gotten that velocity from the inelastic collision with the ball

V = (M1 * U1 + M2 * U2) / (M1 + M2)

0.28 = (85 * 0 + 0.43 * U2) / (85 + 0.43)

U2 = 55.62 ft/s