Projectile motion basketball ball shot help

In summary, the basketball player must shoot the ball with an initial velocity of 7.73m/s in order to make it 10.0m away from the 3.05m high basketball net and hit the ground with a displacement of 1.05m.
  • #1
jamescb82
6
0
Homework Statement
So I am given the height of a basketball player (2.00m tall) and the height of a basketball net (3.05m tall) and the distance at which the player stands from the net (10.0m away). The player shoots the ball at an angle of 45.0 degrees. What must the initial velocity be to make this shot?

(Image of attempt and question attached)

So I am completely lost mostly because of the fact that multiple people have told me many different thing. I was hoping someone could point me in the right direction in understanding the needed formulas. Thank you in advance!
 

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  • #2
Hello and welcome to PF!

We ask that you show some attempt at solving the problem. Looks like you have all the equations that you need (and then some). Based on the information given, which of your equations do you think can be used to get the answer?
 
  • #3
If I numbered these equations 1 - 4, top to bottom, I'm guessing number two and four?
 
  • #4
You might be right. To help you see (without guessing) first take each number given in the problem (I think there are 4) and try to associate that number with one of the symbols in the equations.
 
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  • #5
So I attempted to solve for the final velocity and thought about the horizontal and vertical components of this problem. In the end I took these equations and used substitution to solve for time (1.83s) and used that time to solve for velocity initial (7.73m/s). I don't have the answers to these questions. Can someone help me confirm my solution?
 

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  • #6
If you substitute your answers for vo and t into the equation y = - (1/2)gt2 + (vosinθo)t + yo, do you get the expected result for y?

If you can show the details of your calculation, we can check your work.
 
  • #7
Xf=range (distance from net = 10.0m)
Yf=displacement above initial launch point (3.05m - 2.00m = 1.05m)
VA=initial velocity
 

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  • #8
Overall, your work looks good. But when you solved the equation 1.05 = 10.0 - 4.9t2, did you forget that t is squared?
 
  • #9
Yes you're right! Thank you for catching me on that so t should be 1.35s which brings the final velocity to 10.5 m/s
 
  • #10
Looks right. Good work!
 
  • #11
Thanks for the help!
 

1. What is projectile motion?

Projectile motion is the curved path that an object takes when it is thrown or launched into the air. It is a combination of horizontal and vertical motion.

2. How does projectile motion apply to shooting a basketball?

When shooting a basketball, the ball follows a curved path due to the force of gravity pulling it downward. This is an example of projectile motion.

3. What factors affect the trajectory of a basketball shot?

The trajectory of a basketball shot is affected by several factors, including the initial velocity of the shot, the angle at which the ball is launched, the force of gravity, and air resistance.

4. How can I improve my basketball shot using knowledge of projectile motion?

Understanding projectile motion can help you improve your basketball shot by allowing you to adjust your aim and power to compensate for the effects of gravity and air resistance. It can also help you determine the best angle to shoot from for maximum accuracy.

5. Can you predict the exact trajectory of a basketball shot?

No, it is impossible to predict the exact trajectory of a basketball shot due to the ever-changing variables such as air resistance and minor changes in initial velocity. However, by understanding projectile motion, you can make educated estimations and adjustments to improve your shot.

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