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

In summary, the conversation discusses a physics problem involving a boy throwing a ball upward with a speed of 12m/s while being affected by a horizontal wind acceleration of 0.4m/s^2. The goal is to find the angle at which the ball must be thrown so that it returns to the point of release, assuming the wind does not affect the vertical motion. The conversation includes calculations and equations for both the x and y directions, with two unknowns being the angle and time. The solution is found by substituting the equations for x and y, with x and y both being equal to 0 when the ball returns to the hand.
  • #1
Larrytsai
228
0

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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  • #2
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:
 
  • #3


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.
 
  • #4
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:
 
  • #5


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
 
  • #6
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.
 
  • #7


What substitutions were made?
 

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

1) What is the angle theta baseball problem?

The angle theta baseball problem is a physics problem that involves calculating the optimal angle at which a baseball should be thrown from a pitcher to achieve the longest possible distance. It takes into account factors such as air resistance, gravity, and the initial velocity of the ball.

2) Why is the angle theta important in baseball?

The angle theta is important in baseball because it affects the trajectory and distance of a thrown ball. A higher angle can result in a longer distance, but it also increases the risk of the ball being caught by the opposing team. Therefore, finding the optimal angle is crucial for a successful throw.

3) How is the angle theta calculated?

The angle theta can be calculated using the following formula: θ = tan⁻¹((2gh) / v²), where θ is the angle, g is the acceleration due to gravity, h is the height of the release point, and v is the initial velocity of the ball. This formula takes into account the factors that affect the ball's trajectory and distance.

4) What is the optimal angle theta for maximum distance in baseball?

The optimal angle theta for maximum distance in baseball is approximately 45 degrees. This angle allows the ball to travel the farthest distance while also minimizing the risk of being caught by the opposing team.

5) How does air resistance affect the angle theta in the baseball problem?

Air resistance plays a significant role in the angle theta baseball problem. It causes the ball to slow down, making it fall at a steeper angle. This means that a higher angle theta is needed to achieve the same distance as in a vacuum. Therefore, the optimal angle theta in real-life scenarios may be slightly higher than the calculated 45 degrees.

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