A baseball has a velocity of 40.0 m/s (89.5 mi/h), directed horizontally, as it is released...

A baseball has a velocity of 40.0 m/s (89.5 mi/h), directed horizontally, as it is released by a pitcher. The ball's velocity 0.0100 s before it is released is 38.1 m/s, directed 4.00° above the horizontal. Calculate the ball's instantaneous acceleration just before it is released.

Magnitude?

Direction___ west of south?

Solutions

Expert Solution

Given

velocity of baseball v = 40.0 m/s directed horizontally

before 0.01 s , the ball velocity is 38.1 m/s , directed with an angle 4 degrees with horizontal

writing the vector form of velocity before 0.01 s , is

V = 38.1 cos4 i + 38.1 sin4 j

= 38.007 i + 2.658 j

now change in velocity is = (40-38.007)i + (0-2.658)j

= 1.993 i - 2.658 j

and the acceleration is A = 1.993/0.01 i -2.658 /0.01 j
A=199.3 i - 265.8 j

the magnitude of the acceleration is = sqrt(199.3^2+(-265.8)^2) m/s2 = 332.22 m/s2

the direction is theta = arc tan (-265.8/199.3)

theta = -53.13 degrees
which is west of south.

genius_generous answered 1 year ago

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