When you hit a 0.27kg volleyball the contact time is 50ms and the average force is 125N.
If you serve the volleyball (from rest) sending it up at 30 degrees above the horizontal, what is the vertical component of the momentum of the volleyball in kgm/s ?
To find the vertical component of the momentum of the volleyball, we need to first calculate its initial velocity in the vertical direction.
To do this, we can use the fact that the volleyball is launched at a 30 degree angle above the horizontal, and that the initial velocity in the vertical direction is given by:
vy = v * sin(30)
where v is the initial velocity of the volleyball, and sin(30) is the sine of 30 degrees.
To find v, we can use the fact that the average force during contact was 125N and the contact time was 50ms. We can use the impulse-momentum theorem:
J = Favg * t = m * Δv
where J is the impulse, Favg is the average force, t is the contact time, m is the mass of the volleyball, and Δv is the change in velocity of the volleyball during contact.
Rearranging this equation to solve for Δv, we get:
Δv = J / m = Favg * t / m = 125 * 0.05 / 0.27 = 23.15 m/s
This means that the initial velocity of the volleyball is 23.15 m/s. Plugging this into the equation for vy, we get:
vy = 23.15 * sin(30) = 11.58 m/s
Finally, we can find the vertical component of the momentum by multiplying the mass of the volleyball by the velocity in the vertical direction:
p(y) = m * vy = 0.27 * 11.58 = 3.1254 kgm/s (rounded to 4 decimal places)
Therefore, the vertical component of the momentum of the volleyball is 3.1254 kgm/s.