A major leaguer hits a baseball so that it leaves the bat at a speed of 29.0 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 29.0

m/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?

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

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?

Explanation / Answer

voy = v*sin(36.5) = 17.25 m/s
vox = v*cos(36.5) = 23.3 m/s


A) y = voy*t -0.5*g*t^2


10 = 17.25*t - 4.9*t^2

4.9*t^2 - 17.25*t +10 = 0

solving above eqn we get, t1 = 0.732 s and t2 = 2.789 s

B) it always remains asame

vox = v*cos(36.5) = 23.3 m/s

C)

vy = voy - g*t1 = 17.25 - 9.8*0.732 = 10.078 m/s

vy = voy - g*t2 = 17.25 - 9.8*2.789 = -10.078 m/s

D) v = 29 m/s

E) velocity makes 36.5 degrees below the x axis

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