Find the angles for the basketball throw, projectile problem

  • Thread starter gpmattos
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1. A basketball will be launched of height h and will hit the hoop that has a height of H. The horizontal distance between the ball and the hoop is d. Given an initial velocity Vo write both possible launch angles in function of h, H, Vo and d


2.
theta=tg^-1(Vy/Vx) (1)
Vx(t)=Vo*cos(theta) (2)
Vy(t)=Vo*sen(theta)-g*t (3)
X(t)=0+Vo*cos(theta)*t (4)
Y(t)=0+Vo*sen(theta)*t-g^2/t (5)



The Attempt at a Solution


I'm lost in how to solve this problem. My main question is:

To find the angles i have to substitute both (2) and (3) in equation (1), but when i try to do that i have the "t" in the equation. if i try to use (4) to isolate t (1) will have theta and i can't isolate it. I'm clueless as how to procede.
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi gpmattos! welcome to pf! :smile:
To find the angles i have to substitute both (2) and (3) in equation (1), but when i try to do that i have the "t" in the equation. if i try to use (4) to isolate t (1) will have theta and i can't isolate it. I'm clueless as how to procede.
you should have an x equation and a y equation with the same t (and the same θ)

eliminate t, and that should give you an equation in θ that you can solve …

show us how far you get :smile:
 
  • #3
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I found some information on what i needed to find the angles and i think i did it right, but i have one question, but first i'll explain my approach:

When the ball reaches its maximum height:(t1 seconds have passed)

Vy(t)=0=Vo*senb-gt==> vsenb=gt ==> t1=vsenb/g

also:

Y(t)=H-h=L=g/2(t1^2-t2^2) being t2 the time the ball reaches its destination

X(t)=Vo*cosb*(t1+t2)


We can rewrite the equation for the Y-axis:
L=g/2(t1+t2)*(2t1-(t1+t2))

from the first equation we know that t1=vsenb/g

from the second equation we know that t1+t2=d/Vo*cosb

then: L=g/2*d/Vo*cosb*((2vosenb/g)-(d/vo*cosb))


After a lot of math we obtain:

g*d^2*tg^2b-2*d*Vo^2*tgb+((L*2*vo^2)+(g*d^2))=0

and we can solve this for tgb





My main question here is, why on the second equation do i substitute t for (t1^2-t^2) and not (t1-t2)^2 . I'm sorry if this is too trivial, but i can't seem to figure this out.
 
  • #4
tiny-tim
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hi gpmattos! :smile:

(just got up :zzz:)
… When the ball reaches its maximum height:(t1 seconds have passed) …
(try using the X2 and X2 buttons just above the Reply box :wink:)

that probably works (i haven't checked your equations),

but you don't need to split the path into two …

just use one t, the time (t2) that the ball goes through the hoop :wink:
 

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