A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0m/s and at an angle of 36.5? above the horizontal. You can ignore air resistance.

A) At what two times is the baseball at a height of 10.0m above the point at which it left the bat?

C) Calculate the vertical component of the baseball's velocity at each of the two times you found in part (a).

D) What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?

E) What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

A man stands on the roof of a 19.0m -tall building and throws a rock with a velocity of magnitude 30.0m/s at an angle of 42.0? above the horizontal. You can ignore air resistance.

B) Calculate the magnitude of the velocity of the rock just before it strikes the ground.

C) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Explanation / Answer

a ) 10 = 22.5*t -4.9*t*t

t1 = 0.498 sec
t2 = 4.09 sec

c)
Uy =28sin36.5 = 16.65 m/s

t1
Vy = 11.769 m/s
t2
Vy = -23.432 (means downwards)

d)

magntiude = 28 m/s

e)

36.5 degrees belwo horizontal

2)

magnitude = 35.63 m/s

horizontal distance = 144.03 m

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