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

here,

velocity of ball ,u = 32.8 m/s

the angle theta = 36.8 degree

ux = 32.8 * cos(36.8)

ux = 26.26 m/s

uy = 32.8 * sin(36.8)

uy = 19.65 m/s

part a) t1 = 0.62 s

t2 = 3.39 s

part b)

let the vertical component of the baseball's velocity at t1 = 0.62 s be vy

using first equation of motion

vy = uy - g *t1

vy = 19.65 - 9.8*0.62

vy = 13.57 m/s

the vertical component of the baseball's velocity at an earlier time calculated in part a) is 13.57 m/s

part c)

let the vertical component of the baseball's velocity at t2 = 3.39 s be vy2

vy2 = uy - g *t2

vy2 = 19.65 - 9.8*3.39

vy = - 13.57 m/s

the vertical component of the baseball's velocity at a later time calculated in part (a) is 13.57 m/s downwards

part d)

the magnitude of the baseball's velocity when it returns to the level at which it left the bat is 19.65 m/s

part e)

the direction of the baseball's velocity when it returns to the level at which it left the bat , theta1 = 180 - theta

theta1 = 143.2 degree

the direction of the baseball's velocity when it returns to the level at which it left the bat is 143.2 degree counterclockwise from the horizontal

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