A basketball player jumps straight up to launch a long jump shot atan angle of 4

Question

A basketball player jumps straight up to launch a long jump shot atan angle of 45 degrees with the horizontal and a speed of 15 m/s.The 75 kg player is momentarily at rest at the top of his jump justbefore 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 pointparticle. Assume the ball is launched to the right (in the positivex and y directions).

a. What is the player's velocity immediately after the shot isreleased?
b. How far from his original position does he land? (Answer must berelative to original position)

Explanation / Answer

This problem can be solved using conservation of linear momentum principle initial momentum is equal to final momentum as the total system(player, ball)  was at rest before the shot.total liner momentum initially is zero. velocity of ball initially (Vbi) = velocity of player (Vpi) =0 In x,y directions final velocity of ball given =15m/s angle of launch with horizon=45 velocity of ball in x component (Vbx) =15 * cos(45) = 15/root(2) velocity of ball in y component (Vby) =15 * sin(45) = 15/root(2) Let velocity of player in x component = Vpx velocity of player in y component = Vpy mass of ball =m=0.62kg mass of player =M=75 kg by conservation of l momentum in x directiom M*Vpx + m*Vbx =0 M*Vpx = -m*Vbx Vpx = - m*Vbx/M = - (0.62 * (15/root(2))/75 =- 0.087681241m/s by conservation of l momentum in y directiom M*Vpy + m*Vby =0 M*Vpy = -m*Vby Vpy = - m*Vby/M = - (0.62 * (15/root(2))/75 =- 0.087681241m/s so velocity of player is given by =0.124 m/s immediately after release of ball now comes the laws of motion time taken by a body which has vertical velocity (-Vpy) (down ward) to reach ground is given by s = u*t + (1/2)*g*t*t s is distance travelled by particle = 0.80 m u initial velocity = -Vpy = 0.087681241m/s g acceleration due to gravity = 9.8 m/sec2 by solving the quatriatic equation we willget =0.395212997sec so distance moved by playe from initial position is given by D = Vpx*t =0.395212997*0.087681241 =0.034652766 m.

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