A major leaguer hits a baseball so that it leaves the bat at a speed of 32.8 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 32.8 m/s and at an angle of 36.8 above the horizontal. You can ignore air resistance.

part (a) t1= .620 t2= 3.39

part (b) Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

part (c) Calculate the vertical component of the baseball's velocity at a later time calculated 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

b)

the vertical component = 32.8 * sin(36.8deg) = 19.65 m/s

at earlier time t = 0.62 sec

v = u + at

v = 19.65 - 9.8 * 0.62 = 13.57 m/s

c)

then for the later time

t= 3.39 sec

usng the same formula

v = 19.65 - 9.8 * 3.39 = -13.57 m/s

d)

here the velocity is same as inital = 32.8 m/s

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