A major leaguer hits a baseball so that it leaves the bat at a speed of 30.6 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.6 m/s and at an angle of 34.7 above the horizontal. You can ignore air resistance.

Part A: At what two times is the baseball at a height of 10.5 m above the point at which it left the bat?

Part B: Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part (a).

Answer: 25.2 m/s

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

Answer: 9.88 m/s

Part D: Calculate the horizontal component of the baseball's velocity at a later time calculated in part (a).

Part E: Calculate the vertical component of the baseball's velocity at a later time calculated in part (a).

Part F: What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

v = 30.6 m/s at 34.7 degree above horizontal

intial horizontal velocity = horizontal velocity throughout = 30.6 cos 34.7 = 25.16 m/s

initial horizontal velocity, uv = 30.6 sin 34.7 = 17.42 m/s

A) height, S = 10.5 m

S = uv t - (1/2) g t2

=> 10.5 = 17.42 t - 4.905 t2

=> t = 0.77 s or 2.78 s

So ball is at height 10.5 above impact at 0.77 s and 2.78 s

B) Horizontal velocity at 0.77 = 25.16 m/s (constant at all time)

C) Vertical velocity at 0.77 s = uv - 9.81t = 17.42 - 9.81*0.77 = 9.86 m/s

D) Horizontal velocity at 2.78 = 25.16 m/s (constant at all time)

E) Vertical velocity at 2.78 s = uv - 9.81t = 17.42 - 9.81*2.78 = - 9.86 m/s

F) Velocity at same height remains same ( conservation of energy)

=> velocity when ball returns to initial level = initial velocity = 30.6 m/s

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