? asked in Science & MathematicsPhysics · 1 decade ago

A 2.05 m tall basketball player attempts a goal 10.2 m from a basket (3.05 m high). If he shoots the ball at a?

A 2.05 m tall basketball player attempts a goal 10.2 m from a basket (3.05 m high). If he shoots the ball at a 45° angle, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard?

A projectile question

????m/s

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  • 1 decade ago
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    The question does not state anything about how the player is holding the basketball. It also does not specify whether the player has jumped or if the player did, how high the player jumped.

    I will assume that the height of the basketball player is specified because that is the height above the ground that the player releases the ball. Most shots are not from exactly a players height but there is no other mention of something that could lead one to guess how high the ball was when it left the players hands.

    The problem does not specify what it means for the attempt to be made 10.2 m from the basket. I will assume that the distance is measured across the floor from directly below the center of the ball when the player releases the ball to a point on the floor directly under the center of the ball.

    assume that the ball travels up from 2.05 m above the floor starting at 45 degrees and then curves over and goes through the rim of the net at 3.05 m above the floor.

    The answer is 10.5m/sec

    The choice of 45 degree angle to launch the ball means that the velocity component Parallel to the ground is equal to the velocity component perpendicular to the ground. The component of the velocity parallel to the ground is not significantly different throughout the travel of the ball from hand to hoop. The component of the velocity vertical to the ground diminshes to zero as it reaches it's maximum height and then accelerates to something less than the original absolute value when it goes through the hoop

    let VH be the horizontal velocity at the time that the ball is released.

    The vertical component of the initial velocity is therefore also VH.

    The total time of the travel of the ball in seconds is 10.2/VH

    The time traveling up is VH/g=VH/9.8

    The distance traveling up is 1/2*9.8*VH/9.8*VH/9.8=VH*VH/19.6

    The distance traveling down is (VH*VH/19.6)-1

    The time traveling down is sqrt(2*((VH*VH/19.6)-1)/9.8)

    10.2/VH=VH/9.8+sqrt(2*((VH*VH/19.6)-1)/9.8)

    I am tired and will leave the rest for you to do.

  • Anonymous
    4 years ago

    Let u = initial speed of basketball V0x = u cos 45 V0y = u sin 45 x = Vx(t), Vx=V0x, Vx = u cos 45 9.9 = ut cos 45 t = 9.9 / u cos 45 y = y0 + V0y*t - (0.5)gt^2 1.01 = ((u sin 45) * (9.9 / u cos 45)) - (0.5)(9.8)(9.9 / u cos 45)^2 1.01 = (9.9 tan 45) - (0.5)(9.8)(9.9^2 / (u^2 * (cos 45)^2)) (1.01-(9.9 tan 45))/(-0.5*9.8) = (9.9^2 / (u^2 * (cos 45)^2)) (u^2 * (cos 45)^2)) = 9.9^2 /((1.01-(9.9 tan 45))/(-0.5*9.8)) u^2 = (9.9^2 /((1.01-(9.9 tan 45))/(-0.5*9.8))) / (cos 45)^2 u = 10.394 m/s Yes, I know the simplifying gets complicated as it's hard to show on here. Basically, you follow these steps: A) 9.9 * tan(45) B) 1.01 - A C) B / (-0.5*9.8) D) 9.9^2 / C E) D / (cos(45))^2 F) sqrt(E) = 10.394 m/s = answer!!!

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