Projectile motion (one try left, 2 hours to complete)

[I_LuV_FiZiX]

Homework Statement

When baseball outfielders throw the ball, they usually allow it to take one bounce on the theory that the ball arrives sooner this way. Suppose that after the bounce the ball rebounds at the same angle theta as it had when released (see figure below) but loses half its speed. Assuming the ball is always thrown with the same initial speed, at what angle theta should the ball be thrown in order to go the same distance D with one bounce (lower path) as one thrown upward at alpha = 48.4° with no bounce (upper path)?

******THIS IS NOT WHAT I AM ASKING. I calculated this to be the correct answer of 26.3 degrees; this was only included because it contains information needed to answer the question I am unsure of.

WHAT I AM ASKING: Determine the ratio of the times for the one-bounce and no-bounce throws.

Homework Equations

Any of the time equations for projectile motion (unsure what they are)

Dx = Vx(t)

The Attempt at a Solution

I used the above equation and set it up so that the ratio would equal the time of the first divided by the time of the second. Many variables canceled out here (V's and D's), and I was left an expression ratio = 3cos26.3/cos48.4..this gave me a ratio of 4.05..The computer said this was incorrect. I was told by my classmates that because the question was not exactly clear as to the ratio of what to what, the inverse of the value calculated was correct for some of them. I only have one try left, and am too scared to put it in because I would like to make sure that my derivation of this time ratio equation is correct. Any help would be GREATLY appreciated

 

[Hootenanny]

Don't forget that for the path where the ball bounces, the velocity of the ball is not constant.

 

[baba89]

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I would first clarify the question with the teacher or professor to ensure that I fully understand what is being asked. From my understanding, the question is asking for the ratio of the times it takes for a one-bounce throw and a no-bounce throw to travel the same distance D.

To find this ratio, we can use the equation for time in projectile motion, t = 2Vsin(theta)/g, where V is the initial velocity, theta is the angle of release, and g is the acceleration due to gravity. For the one-bounce throw, we can use the given information that the ball rebounds at the same angle theta and loses half its speed, so the initial velocity after the bounce is V/2.

Therefore, for the one-bounce throw, the time it takes to travel distance D can be calculated as t1 = 2(V/2)sin(theta)/g = Vsin(theta)/g.

For the no-bounce throw, we are given an angle of release, alpha = 48.4°, but we do not have the initial velocity. However, since we know that both throws have the same initial speed, we can use the fact that the horizontal distance traveled, Dx, is the same for both throws. Therefore, we can set up the equation Dx = Vx(t) and solve for V, which gives us V = Dx/t. Substituting this into the equation for time, we get t2 = 2(Dx/t)sin(alpha)/g.

Now, to find the ratio of the times, we can divide t1 by t2:

t1/t2 = (Vsin(theta)/g) / (2(Dx/t)sin(alpha)/g)

= (Vsin(theta)/g) * (t/2Dx) * (g/sin(alpha))

= (Vsin(theta)/g) * (t/g) * (1/2Dx) * (g/sin(alpha))

= (Vsin(theta)/g) * (t/g) * (1/2Dx) * (1/sin(alpha))

= (Vsin(theta)/g) * (t/g) * (1/2Dx) * (1/cos(alpha)/cos(alpha))

= (Vsin(theta)/g) * (t/g) * (1/2Dx) * (1/cos(alpha)) * (cos(theta)/cos(alpha))