# Homework Help: Finding Initial Velocity w/o Time to clear a specific height

1. Sep 21, 2009

### jkcali2ooo

1. The problem statement, all variables and given/known data
A basketball player shoots the ball at a height of 1.7m at an angle of 60 degrees into a hoop. The range is 10ft or 3.05m. The height from the ground to the top of the hoop is 10ft or 3.05m. What is the initial velocity m/s to make it to the hoop?

2. Relevant equations

1) Vy^2 = Viy^2 - 2g(Change in Y)
2) Y = Yi + Viy(t) - 1/2gt^2
3) Vy = Viy - gt
4) X = Xi + Vixt = (ViCos(theta)) t

3. The attempt at a solution

B/c the ball is shot at 1.7m off the ground I reduced this to zero thereby reducing the height of the hoop by 1.7m.

So I have the following givens:
x= 3.05m
y = 3.05m - 1.7m = 1.35m
Theta = 60 degrees
Xi = 0 m
Yi = 0 m
Time is unknown
Solving for Initial Velocity (Vi)
I am assuming max height is enough to clear the hoop at 1.35m. So at max height Velocity ='s 0 m/s.

So now I used X = Xi + Vixt = ViCos(Theta)t

3.05 = 0 + Vi (Cos 60) t
t = 6.1 / Vi

Now I take t and plug into y = Yi + Viyt - 1/2gt^2
1.35m = 0 + Visin(60)t -1/2(9.8)(t^2)
1.35 = (0.866025Vi)(6.1/Vi) - 4.9 (6.1/Vi)^2
1.35 = 5.282754 - 182.329/Vi^2
-3.932754 = -182.329 / Vi^2
Vi^2 = 46.36165903
Vi = 6.8089 m / s

Does that seem right or am I not taking into account other factors. Thanks.

Last edited: Sep 21, 2009
2. Sep 21, 2009

### Staff: Mentor

I'm confused by this. Don't you mean to say Viy is 0, not Vmax?

3. Sep 21, 2009

### jkcali2ooo

I believe that intial velocity is unknown since the ball is already traveling at some speed in m/s. However, at the max height of the ball when it reaches the hoop would have a velocity of 0 with a negative accelleration of gravity or -9.8m/s^2.

I have revised my attempt at the problem above. If anyone can confirm or discount my work would be much appreciated.

Thanks!

4. Sep 21, 2009

### Staff: Mentor

Ah, I understand now. You mean Vy(Ymax) = 0. Got it.

5. Sep 21, 2009

### Staff: Mentor

This step looks wrong:

Did you drop the Vi term? The first line is a quadratic equation with both Vi and Vi^2 terms. You would use the quadratic formula to find the solution...

6. Sep 21, 2009

### jkcali2ooo

Sorry I added another Vi.

So
1.35 = (0.866025Vi)(6.1/Vi) - 4.9(6.1/Vi)^2
which then leads to...
1.35 = 5.282754 - 182.329 / Vi^2

So the Vi's cancel eachother out so I don't have to use the quadratic forumula. I wasn't sure about this step either cause it seemed weird that they cancelled out without having to use the quadratic formula to solve for Vi.

Thanks.

7. Sep 21, 2009

### Staff: Mentor

Ah, got it. If you have a graphing calculator, you could plot the trajectory that you've calculated, to see if it does what you expect it to.

8. Sep 21, 2009

### jkcali2ooo

Thanks for the help Berkeman. When I calculate out time and re-plug that back in with the inital velocity I calculated, I would get to the 1.35m (a.k.a. the hoop.)

However, I was wondering if I can just reduce the height where the ball was initially shot from and then just subtract that out of the initial height of the hoop. Is that fine to do w/o changing any other value as far as giving me the same answer? Thanks.

9. Sep 21, 2009

### Staff: Mentor

I believe that is fine. All you are doing is setting the zero point for y to be the start of the ball's flight. You are drawing the origin of your x-y axes at the player's hand instead of at his feet. Good simplification.