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The momentum of the ball when it hits the ground is 4.89 kg.m/s.
The given parameters;
- <em>mass of the baseball, m = 0.145 kg</em>
- <em>height of fall of the ball, h = 58 m</em>
The final velocity of the ball when it hits the ground is calculated as follows;
The momentum of the ball when it hits the ground is calculated as follows;
P = mv
P = 0.145 x 33.72
P = 4.89 kg.m/s
Thus, the momentum of the ball when it hits the ground is 4.89 kg.m/s.
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<h2>Angular acceleration is 80 rad/s²
</h2><h2>Number of revolutions undergone is 1.02</h2>
Explanation:
We have equation of motion v = u + at
Initial angular velocity, u = 0 rad/s
Final angular velocity, v = 32 rad/s
Time, t = 0.40 s
Substituting
v = u + at
32 = 0 + a x 0.40
a = 80 rad/s²
Angular acceleration is 80 rad/s²
We have equation of motion s = ut + 0.5 at²
Initial angular velocity, u = 0 rad/s
Angular acceleration, a = 80 rad/s²
Time, t = 0.4 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.4 + 0.5 x 80 x 0.4²
s = 6.4 rad
Angular displacement = 6.4 rad
Number of revolutions undergone is 1.02