A baseball has a velocity of 40.6 m/s (90.8 mi/h), directed horizontally, as it is released by a pitcher. The ball's velocity 0.0100 s before it is released is 38.7 m/s, directed 3.75° above the horizontal. Calculate the ball's instantaneous acceleration just before it is released.
magnitude | |
direction | below the horizon |
If the horizontal direction is assumed to be along the x axis
and the vertically up direction is along the +y axis, then, the
initial velocity is
Now as
,
, so,
And the final velocity is
The elapsed time is
And so, the acceleration is
The magnitude of the acceleration
And the angle it makes with the horizontal is
The negative sign implies the angle is below the horizontal. And
so, the magnitude of the angle below the horizontal is
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