A ball is thrown upward. It leaves the hand with a velocity of 13.8 m/s, having been accelerated through a distance of 0.460 m. Compute the ball's upward acceleration, assuming it to be constant.
here,
the final velocity , v = 13.8 m/s
s = 0.46 m
let the accelration be a
initial velocity , v0 = 0 m/s
using third equation of motion
v^2 - v0^2 = 2 * a * s
13.8^2 - 0 = 2 * a * 0.46
solving for a
a = 207 m/s^2
the accelration is 207 m/s^2
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