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 34.9 degree above the horizontal. You can ignore air resistance.

A. At what two times is the baseball at a height of 9.00m above the point at which it left the bat?
(Please enter your answer as two numbers, separated with a comma, in order t1, t2 where t2>t1.)

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

Vo = 31m/s
Xo = 31*cos34.9 = 25.42 m/s.
Yo = 31*sin34.9 = 17.7365 m/s.

a. h = Yo*t + 0.5g*t^2 = 9 m
17.7365*t - 4.9t^2 = 9
-4.9t^2 +17.7365t - 9 = 0. Use Quad. Formula.
t = 0.610342 sec,3.01 sec

b. X = Xo = 25.42 m/s. = Hor. component
of velocity , and does not change.

c. Y = Yo + g*t = 17.7365 - 9.8*0.61 = 11.75 m/s.
Y = 17.7365 - 9.8*3 = -11.75 m/s.

d. V = Vo = 31m/s

e. 34.9 degree below the horizontal.
downward

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