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 17.2 inches off the ground. He hits the ground 1.61 inches away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball travelling? I am not sure where to begin on this, I wish to learn how to do this problem, please help. Thank you.
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The time needed to free fall a height of 1.61 in (=1.433 ft) is:
s = a×t²/2
1.433 = 32×t²/2
t = 0.3 s
In that time, he makes a horizontal displacement of 1.61 in (=0.1342 ft).
So his horizontal velocity was:
r = v × t
0.1342 = v × 0.3
v = 0.4473 ft/s
He must have gotten that velocity from the inelastic collision with the ball.
V = (M₁×U₁ + M₂×U₂) / (M₁ + M₂)
0.4473 = (85×0 + 0.43×U₂) / (85 + 0.43)
U₂ = 88.87 ft/s < - - - - - - velocity of the ball when caught
s = a×t²/2
1.433 = 32×t²/2
t = 0.3 s
In that time, he makes a horizontal displacement of 1.61 in (=0.1342 ft).
So his horizontal velocity was:
r = v × t
0.1342 = v × 0.3
v = 0.4473 ft/s
He must have gotten that velocity from the inelastic collision with the ball.
V = (M₁×U₁ + M₂×U₂) / (M₁ + M₂)
0.4473 = (85×0 + 0.43×U₂) / (85 + 0.43)
U₂ = 88.87 ft/s < - - - - - - velocity of the ball when caught