What Angle Should a Basketball Player Aim to Score from 15 Feet Away?

[negation]

Homework Statement

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?

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°

 

[BvU]

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

 

[negation]

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

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

 

[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 !

 

[negation]

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 !

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

 

[BvU]

vi/vi = 1

 

[negation]

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 !

The answer doesn't tally with the book's

 

[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...

 

[haruspex]

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...

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.

 

[negation]

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...

I overlooked that but still it's not working out

 

[negation]

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.

Let me try

 

[negation]

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.

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

 

[haruspex]

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

I get a much larger angle. Please post your working.

 

[negation]

i get a much larger angle. Please post your working.

 

[haruspex]

View attachment 66141

6th line (1.8 v_i² = ...), check the sign on the tan² term.

 

[negation]

6th line (1.8 v_i² = ...), check the sign on the tan² term.

Ought to have been a negative

 

[negation]

6th line (1.8 v_i² = ...), check the sign on the tan² term.

Did you got 72.8?

 

[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)

 

[negation]

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)

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.

 

[negation]

solved.

 

[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.