Part A Constants Find the magnitude of the initial velocity of the baseball (the

Question

Part A Constants Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown). A baseball thrown at an angle of 55.0° above the horizontal strikes a building 17.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance U E m/s Submit Request Answer Part B Find the magnitude of the velocity of the baseball just before it strikes the building m/s Submit Request Answer Part C Find the direction of the velocity of the baseball just before it strikes the building below the horizontal

Explanation / Answer

a)

Initial Horizontal and vertical components of velocities are

Vox=VoCos55

Voy=VoSIn55

Horizontal distance is

X=Voxt

=>17=(VoCos55)t

t=(17/VoCos55)

From

Y=Yo+Voyt-(1/2)gt2

8=0+(VoSin55)(17/VoCos55) -(1/2)*9.8*(17/VoCos55)2

8=17tan55-4304.4/Vo2

Vo =16.26 m/s

b)

TIme taken by baseball to strike the building is

t=17/(16.26Cos55)

t=1.823 s

From

Vfy=Voy-gt

Vfy =(16.26Sin55)-9.81*1.823

Vfy=-4.562 m/s

Vfx=Vox=16.26Cos55=9.326 m/s

Magnitude

V=sqrt[9.3262+(-4.562)2]

V=10.382 m/s

c)

Direction

o=tan-1)(-4.562/9.326)=-26.07o

o=26.07o below the horizontal

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