Projectile Motion Free Throw?

  • Thread starter Devin Humphreys
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  • #1
Devin Humphreys

Homework Statement


A basketball player practices shooting three-pointers from a distance of 7.71 m from the hoop, releasing the ball at a height of 1.93 m above ground. A standard basketball hoop's rim top is 3.05 m above the floor. The player shoots the ball at an angle of 45° with the horizontal. At what initial speed must he shoot to make the basket?

V0 = _______________ m/s

Homework Equations



The four kinematic equations
Parabolic trajectory equation (simplified since x0 is taken to be 0 to

y = y0 + ((vy0 / vx0) * x) - ((g / (2vx02) * x2)




The Attempt at a Solution



I know that the way the problem is set up it will take the same amount of time for the basketball to travel 7.71 meters horizontally as to reach the peak of its trajectory then go down to 3.05 meters in the y-direction for a displacement of 1.12 m. I don't know how to get from there to either finding the velocity directly or finding the time. If I can find the time, I can plug it and the y-displacement into the kinematic equation y = yi * t + (1/2)gt2 to get the initial velocity in the y direction, which I can apply basic trig to in order to find the total velocity. I just don't know how to find the time.
 

Answers and Replies

  • #2
haruspex
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Homework Statement


A basketball player practices shooting three-pointers from a distance of 7.71 m from the hoop, releasing the ball at a height of 1.93 m above ground. A standard basketball hoop's rim top is 3.05 m above the floor. The player shoots the ball at an angle of 45° with the horizontal. At what initial speed must he shoot to make the basket?

V0 = _______________ m/s

Homework Equations



The four kinematic equations
Parabolic trajectory equation (simplified since x0 is taken to be 0 to

y = y0 + ((vy0 / vx0) * x) - ((g / (2vx02) * x2)




The Attempt at a Solution



I know that the way the problem is set up it will take the same amount of time for the basketball to travel 7.71 meters horizontally as to reach the peak of its trajectory then go down to 3.05 meters in the y-direction for a displacement of 1.12 m. I don't know how to get from there to either finding the velocity directly or finding the time. If I can find the time, I can plug it and the y-displacement into the kinematic equation y = yi * t + (1/2)gt2 to get the initial velocity in the y direction, which I can apply basic trig to in order to find the total velocity. I just don't know how to find the time.
You have two equations, one for horizontal motion and one for vertical. You have two unknowns, the initial angle and the time.
Two equations, two unknowns... solve.
 
  • #3
Devin Humphreys
You have two equations, one for horizontal motion and one for vertical. You have two unknowns, the initial angle and the time.
Two equations, two unknowns... solve.

I have the initial angle (that's 45 degrees). What I don't have is the initial velocity, and that significantly complicates things. Is there some way I can use the initial angle to compensate even if both variables are found in both equations?
 
  • #4
haruspex
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I have the initial angle (that's 45 degrees). What I don't have is the initial velocity, and that significantly complicates things. Is there some way I can use the initial angle to compensate even if both variables are found in both equations?
Sorry, I meant initial speed, not angle. But it's the same deal. Let the time be t and the initial speed be v. What two equations can you write?
 

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