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A baseball has a velocity of 40.9 m/s (91.5 mi/h), directed horizontally, as it is released...

A baseball has a velocity of 40.9 m/s (91.5 mi/h), directed horizontally, as it is released by a pitcher. The ball's velocity 0.0100 s before it is released is 38.3 m/s, directed 2.25° above the horizontal. Calculate the ball's instantaneous acceleration just before it is released. magnitude ? direction below the horizon ?

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Answer #1

Given is:-

u1-38.3m s, 0- 2.25

thus

u138.3cos02.25
38.27m/S

similarly

uly 38.3sin(2.25)
uly-1.504m/s

or

T1- (38.27i 1.504j)m/s

and

(40.91)m/s

Time taken during this change in speed is   t 0.0100s

Now,

change in the velocity Delta u = u_2 - u_1 = (40.9i)-(38.27i +1.504j)

which gives us

Delta u = (2.63 hat i - 1.504 hat j)

thus the acceleration will be

a = rac{Delta u}{Delta t}

by plugging all the values we get

2.63i 1.504 0.01

or

2632-150.4 j

thus the magnitude will be

a(263) (-150.4)2

which gives us

oxed{| a | approx 303 m/s^2}

and the direction will be

heta = tan^{-1}(rac{y}{x})

tan-1 (-150.4 θ

heta = -29.8 ^circ

or

θ-29.8 below the horizon

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