A major leaguer hits a baseball so that it leaves the bat at a speed of 31.0 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 31.0 m/s and at an angle of36.9degrees above the horizontal. You can ignore air resistance.

A) At what two times is the baseball at a height of 9.50 m above the point at which it left the bat?

Enter your answer as two numbers, separated by a comma, in the 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.

Enter your answer as two numbers, separated by a comma, in the order v1, v2.

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

Enter your answer as two numbers, separated by a comma, in the order v1, v2.

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

vi=31.0m/s and =36.9deg

A) Use equation

h=viy*t+1/2at^2

9.5=31sin36.9*t+1/2(-9.8)t^2

t1=0.6s anr t=3.2s

B) Use equation

vfx=vix+ax*t

Since there is not acceleration along x axis ax=0m/s^2

So vfx=vix=vicos = 31.0cos36.9=

v1 =21.62m/s ,   v2=21.62m/s

C) Use equation

vfy=viy+ay*t

ay= g=-9.8m/s^2

at t=0.6s

vfy=31.0sin36.9 - 9.8*0.6 = 12.73 m/s

at t=3.2s

vfy=31.0sin36.9 - 9.8*3.2 = -12.75 m/s

Thus v1=12.73m/s and v2= -12.75 m/s

D) By conservation of energy it will have same magnitude as it left the bat

So vf=31.0m/s

E) Now angle is 36.9 deg but below the horizontal

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