A basketball player jumps straight up to launch a long jump shot at an angle of

Question

A basketball player jumps straight up to launch a long jump shot at an angle of 450 with the horizontal and a speed of 15 m/s. The 75-kg player is momentarily at rest at the top of his jump just before the shot is released, with his feet 0.80 m above the floor. The mass of a basketball is 0.62 kg. Treat the player as a point particle. Assume that the ball is launched to the right (in the positive x and y directions). (a) What is the player?s velocity immediately after the shot is released? (b) How far from his original position does he land? Your answer should be relative to the original position.

Explanation / Answer

The player is momentarily at rest at the peak of his jump.      
      
Thus, the total momentum along x = 0, which is also equal to the final momentum:      
      
0 = m1v1fx + m2v2fx      
      
As m2 = 0.62 kg, v2fx = 15cos45 m/s, m1 = 75 kg, then      
      
v1fx =    -0.088   m/s
      
Now, for the sum along y,      
      
0 = m1v1fy + m2v2fy      
      
As v2fy = 15sin45 m/s,      
      
v1fy =    -0.088   m/s
      

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Thus, the answer to part A is (-0.088) m/s i^ + (-0.088) m/s j^.   [PART A ANSWER]

[Please fill both blanks with -0.088]

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Now, as he is 0.80 m above the ground, the time it takes for him to hit the ground is given by      
      
delta(y) = voy t - 1/2 gt^2      
      
As voy = -0.088, delta(y) = -0.80, then      
      
t =    0.3952   s
      
Thus, at that point, he would have travelled x = vox t horizontally,      
      
x =    0.035   m   [ANSWER, PART B]

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