A baseball is thrown from the roof of h = 20.0 m -tall building with an initial

Question

A baseball is thrown from the roof of h = 20.0 m -tall building with an initial velocity of magnitude 11.1 m/s and directed at an angle of 53.1 above the horizontal.

1/ What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.

2/ What is the answer for part (A) if the initial velocity is at an angle of 53.1 below the horizontal?

3/ If the effects of air resistance are included, will part (A) or (B) give the higher speed?

         a) The part (A) will give the higher speed.

         b) The part (B) will give the higher speed.

Explanation / Answer

Equating energy, we get

1/2mvo2 + mgy = 1/2mv2

cancel m on both sides, we get

v = sqrt (vo2 + 2gy)

v = sqrt (11.12 + 2*9.8*20)

v = 22.698 m/s

(b) As the initial velocity is same,

v = 22.698 m/s

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