A major leaguer hits a baseball so that it leaves the bat at a speed of 31.0m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 31.0m/s and at an angle of 35.4^circ above the horizontal. You can ignore air resistance.

a) At what two times is the baseball at a height of 11.4m 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? (in degrees below the horizontal)

Explanation / Answer

a) h = Vy x t - gt2 /2

11.4 = 31sin35.4 X t - 9.81X t2 /2

11.4 = 17.96t - 4.9t2

t = 0.817 sec & 2.85 sec

b) horizontal component doesn't change with time.

Vx = 31cos35.4 = 25.27 m/s

c) at t = 0.817 sec

Vy = Uy - gt

Vy = 17.96 - 9.81 x 0.817 = 9.95 m/s

d) Vx = 31cos35.4 = 25.27 m/s

e) Vy = 17.96 - 9.81 x 2.85 = 10 m/s downwards

f) at same level speed will be same on direction will change .

V = 31.0 m/s at an angle of 35.4 degres downwards to horizontal

g) 35.4 degrees below the horizontal

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