In: Physics
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?
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.