A major leaguer hits a baseball so that it leaves the bat at a speed of 32.1m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 32.1m/s and at an angle of 34.9(degrees) above the horizontal. You can ignore air resistance (A)At what two times is the baseball at a height of 10.1m above the point at which it left the bat? (B)Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part (a). (C)Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).(D)Calculate the horizontal component of the baseball's velocity at a later time calculated in part (a).(E)Calculate the vertical component of the baseball's velocity at a later time calculated in part (a).(F)What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?(G)What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

horizontal component of velocity 30.2Cos34.0 = 25.03m/s (this remains constant throughout as there is no acceleration in horizontal direction)

vertical component of velocity 30.2Sin34.0 = 16.88m/s

considering only vertical direction and using

s= ut - 1/2gt2

10.6 = 16.88t -1/2*9.8*t2

4.9t2 -16.88t + 10.6 = 0

t= 2.61s and .82s

using v = u -gt

vertical component of velocity at .82s

= 16.88 - 9.8*.82

=8.84m/s

at 2.61

= 16.88 - 9.8*2.61

= -8.69m/sec

when it returns to earth the magnitude is same because there is no change in kinetic energy of the ball as the gravitational field is conservative

when it returns the angle will be -34°

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