A major leaguer hits a baseball so that it leaves the bat at a speed of 29.0 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 29.0 m/s and at an angle of 34.9 above the horizontal. You can ignore air resistance. Part A:At what two times is the baseball at a height of 8.50 m above the point at which it left the bat? Part B: Calculate the horizontal component of the baseball's velocity at each of the two times you found in part A. Part C: Calculate the vertical component of the baseball's velocity at each of the two times you found in part A. Part D: What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat? Part E: What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

Hi,

This problem can be classified as a problem of parabolic movement, where the only force that is present is the weight. Therefore, the only acceleration present in this problem is the one due to gravity, so:

y axis : ay = -g ; x axis : 0

The velocity on each axis would be:

Vox = Vx = cos(34.9°) V = 0.820*29 m/s = 23.8 m/s

Voy = sin(34.9°) V = 0.572*29 m/s = 16.6 m/s

A) The distance in the y axis in terms of time can be calculated as follows:

y = Voy*t + (1/2)at2 ; this give us two solutions for t, which are the two times we are interested in:

8.5 m = 16.6 m/s t - 4.9 m/s2 t2:::::::: 4.9 t2-16.6 t +8.5 = 0 ::::::::: t1 = 0.63 s ; t2 = 2.76 s

B) The movement in the x axis is not affected by any force, so it does not present an acceleration, therefore the speed in this axis does not depends on the time since it remains constant.

V1x = V2x = Vox = 23.8 m/s

C) The velocity in the y axis, in terms of time can be expressed as follows:

Vy = Voy + at

Vy1 = (16.6 m/s) + (-9.8 m/s2)*(0.63 s) = 10.43 m/s ::::: Vy2 = (16.6 m/s) + (-9.8 m/s2)*(2.76 s) = -10.45 m/s

Note: the velocity in each case should have had the same magnitude, but due to certain approximations done when calculating the times, the values are slightly different.

D) The magnitude of the velocity when it returns to the level at which it left the bat should be the same as the magnitude it had when it left said bat:

abs(Vy) = 16.6 m/s

Note: this can be proven using the equation y = Voy*t + (1/2)at2 and then fixing as condition y = 0 , after that, the equation Vy = Voy + at can be used to find the velocity at those times.

E) The direction of the velocity at this instant will be the same it had when it left the bat but inverted, this means it will have an angle of 34.9 below the horizontal.

I hope it helps.

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