# Homework Help: Basketball player shooting

1. Jan 29, 2014

### negation

1. The problem statement, all variables and given/known data

A basketball player is 15ft horizontally from the centre of the basket which is 10ft off the ground At what angle should the player aim the ball from a height of 8.2ft with a speed of 26fts^-1?

3. The attempt at a solution

$x = vi cos . t$
$t = \frac{x}{vi cos Θ}$

sub t into y(t):

1.8 = [vi^2 sin2Θ - gx^2]/2vi^2 cos^2 Θ

Θ= 22.6

The answer stated solution to be 66°

2. Jan 29, 2014

### BvU

Hello again. Don't recognize your y(t) expression. Where did you start from and how did you derive it ?

3. Jan 29, 2014

### negation

Y(t) is the y displacement as a function of time.
It has the form yf = yi + vyi. t - 0.5gt^2
t= x/ vi cos (theta), vi=26ft/s, yf=10, yi=8.2 and g = 32ft/s^2
I sub t and the known variables into the above equation. What I got was 22 degrees. But this contrasts with the book's

4. Jan 29, 2014

### BvU

Funny, I substitute t = x/(vi cos(theta)) in vyi * t with vyi = vi sin(theta) and get x * sin(theta)/cos(theta) !? You can change to 2theta which should introduce a 2 (which you do take into account) but it shouldn't let the x disappear !

5. Jan 29, 2014

### negation

Where is your vi? Shouldn't it be x.vi sinΘ/vi cos Θ?

6. Jan 29, 2014

### BvU

vi/vi = 1

7. Jan 29, 2014

### negation

The answer doesn't tally with the book's

8. Jan 29, 2014

### BvU

I see you multiply by vi^2cos^2(theta) on the righthand side. To keep the equality valid, you have to do so also on the lefthand side. I.e. the 1.8 changes...

9. Jan 29, 2014

### haruspex

Quite so, but I don't think this approach is leading anywhere.
negation, go back to the third line of your attachment: "1.8 = " etc.
sin/cos = tan; 1/cos2 = sec2.
Using sec2 = 1 + tan2 you get a quadratic in tan.

10. Jan 29, 2014

### negation

I overlooked that but still it's not working out

11. Jan 29, 2014

### negation

Let me try

12. Jan 29, 2014

### negation

I'm getting 22.4 degrees. Still, it doesn't tally with the book's

13. Jan 29, 2014

### haruspex

14. Jan 30, 2014

### negation

15. Jan 30, 2014

### haruspex

6th line (1.8 vi2 = ...), check the sign on the tan2 term.

16. Jan 30, 2014

### negation

Ought to have been a negative

17. Jan 30, 2014

### negation

Did you got 72.8?

18. Jan 30, 2014

### BvU

No, should be 65.67 degrees if g = 32
Filling in 72.8 doesn't satisfy your final equation with the corrected sign: some 9630 off!
A check you should always do (if you have time enough)

19. Jan 30, 2014

### negation

Strange. I'm not getting the answer even after correcting that positive sign and changing it to a negative.
But I think it's good enough to leave it here since the problem solving part is much more important than the minute calculation details.

20. Jan 30, 2014

### negation

solved.

21. Jan 30, 2014

### BvU

Congrats! I heartily agree with your #19 post but have to admit that you get more reward, satisfaction, etc. from getting the right answer. And it contributes to saving the world from collapsing bridges and crashing airplanes ;-)
Keep up the good work.