# Projectile motion help

1. Mar 3, 2015

### jamescb82

The problem statement, all variables and given/known data
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. Mar 3, 2015

### TSny

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. Mar 3, 2015

### jamescb82

If I numbered these equations 1 - 4, top to bottom, I'm guessing number two and four?

4. Mar 3, 2015

### TSny

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.

Last edited: Mar 3, 2015
5. Mar 3, 2015

### jamescb82

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. Mar 3, 2015

### TSny

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. Mar 3, 2015

### jamescb82

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. Mar 3, 2015

### TSny

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. Mar 3, 2015

### jamescb82

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. Mar 3, 2015

### TSny

Looks right. Good work!

11. Mar 3, 2015

### jamescb82

Thanks for the help!