Finding the Optimal Angle for a Baseball to Return to its Release Point

AI Thread Summary
The discussion revolves around determining the optimal angle for a baseball thrown upward with an initial speed of 12 m/s, considering a horizontal wind acceleration of 0.4 m/s². Participants are working through the equations of motion in both the x and y directions to find the angle θ that allows the ball to return to its release point. Confusion arises regarding the conditions when the ball lands back in the thrower's hand, specifically that both horizontal and vertical positions (x and y) are zero at that moment. The importance of using the same time variable in both equations is emphasized to solve for the unknowns. Ultimately, the thread highlights the need for careful substitution and understanding of the motion equations to arrive at the correct angle.
Larrytsai
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Homework Statement


A boy throws a ball upward with a speed of vo= 12m/s. The wind imparts a horizontal acceleration of 0.4m/s^2 to the left. At what angle theta must the ball be thrown so that it returns to the point of release. (assuming wind does not affect vertical motion.


Homework Equations


All physics equation


The Attempt at a Solution


I have written down all the formulas in the x and y directions, but it doesn't work out. I have also tried to look for what distance Vy=0 but i still end up having 2 unknowns.
 
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Hi Larrytsai! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)

You should have only two unknowns, θ and t, so two equations should be enough.

Show us your full calculations, and then we can see what went wrong, and we'll know how to help! :smile:
 


ok so what i have so far is,
for y:
y= (1/2 a t^2) + (Vo cosθ t)
(Vcosθ)^2 = (Vocosθ)^2 + 2ay

for x:
x= (1/2 a t^2) + (Vo sinθ t)
(Vsinθ)^2 = (Vosinθ)^2 + 2ax

so far how i see it is I am going to substitute my 2 x equations and my 2 y equations, then ill have 2 equations left to subtitute which are θ and time.
 
Larrytsai said:
ok so what i have so far is,
for y:
y= (1/2 a t^2) + (Vo cosθ t)
(Vcosθ)^2 = (Vocosθ)^2 + 2ay

for x:
x= (1/2 a t^2) + (Vo sinθ t)
(Vsinθ)^2 = (Vosinθ)^2 + 2ax

so far how i see it is I am going to substitute my 2 x equations and my 2 y equations, then ill have 2 equations left to subtitute which are θ and time.

(hmm … the second and fourth equations are just V = V0 … that doesn't help much)

What's the difficulty?

Put all the numbers in, including x = y = 0, and what do you get? :smile:
 


ohh ok, i got the answer, thnx so much, but I am a little confused as to why x is 0, I am thinking that when the height and horizontal distance is 0 , the ball has landed back in the boys hand, therefore the time elapsed is the same so we can subsitute for time?

btw thanks alot
 
Hi Larrytsai! :smile:

Yes, you're getting confused.

When the ball returns to the hand, both x and y are 0, and the time elapsed is t (say).

So the same t appears in both the equation for x and the equation for y.
 


What substitutions were made?
 
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