a .330 kg volleyball is dropped from rest. it takes it 1.30 s to reach the ground. what is the magnitude of its momentum just before it hits the ground

Weight * Time = final momentum

= 0.33 kg*9.8 m/s^2 *1.3 s
= 4.2 kg*m/s

Well, the volleyball certainly had some serious momentum going for it before it hit the ground! To find the magnitude of its momentum, we can use the formula for momentum: p = mv.

Given that the volleyball's mass (m) is 0.330 kg, we just need to figure out its velocity (v) right before it smacks the ground.

Since the volleyball was dropped from rest, we can use the equation of motion d = 1/2 * a * t^2 to solve for acceleration (a). Here, d is the distance traveled (which is the height the volleyball fell), and t is the time it took to fall.

So, if it took 1.30 s for the volleyball to reach the ground, we need to first find the distance it fell. Unfortunately, you didn't mention the height, so I can't calculate the precise value. However, assuming it fell from a typical volleyball height of a few meters, I can help you with a general calculation.

Now, using the distance (d) and time (t) values, we can find the acceleration (a). With the acceleration in hand, we can calculate the velocity (v) using the equation v = a * t.

Once we have the velocity, we can finally calculate the momentum (p) using the formula p = mv. Just multiply the mass (m) by the velocity (v), ensuring that you take the magnitude by ignoring any negative signs.

So, while I can't provide you with an exact answer without the height, I hope this guide helps you solve the problem and calculate that volleyball's impressive momentum!

To find the magnitude of the momentum just before the volleyball hits the ground, we can use the equation:

Momentum = mass × velocity

Given:
Mass of the volleyball (m) = 0.330 kg
Time taken (t) = 1.30 s

To find the velocity, we can use the equation for free-fall distance:

Distance = (1/2) × acceleration × time^2

Considering that the volleyball is dropped from rest, its initial velocity is 0.

Thus, the distance fallen is given by:

Distance = (1/2) × acceleration × time^2

Rearranging the equation:

acceleration = (2 × distance) / (time^2)

Since the volleyball is falling near the surface of the Earth, we can use the acceleration due to gravity (g) as the value for acceleration. On Earth, the average value of g is approximately 9.8 m/s^2.

Therefore:

acceleration = g = 9.8 m/s^2

Let's calculate the distance fallen first:

Distance = (1/2) × acceleration × time^2
= (1/2) × 9.8 m/s^2 × (1.30 s)^2
= 0.5 × 9.8 m/s^2 × 1.69 s^2
= 8.084 m

Now, let's calculate the velocity just before hitting the ground. We can use the equation:

Velocity = acceleration × time

Velocity = g × t
= 9.8 m/s^2 × 1.30 s
= 12.74 m/s

Finally, we can find the magnitude of the momentum:

Momentum = mass × velocity
= 0.330 kg × 12.74 m/s
= 4.2052 kg·m/s

Therefore, the magnitude of the momentum just before the volleyball hits the ground is approximately 4.2052 kg·m/s.

To find the magnitude of momentum just before the volleyball hits the ground, we need to use the formula:

Momentum (p) = mass (m) × velocity (v)

In this case, the mass of the volleyball is given as 0.330 kg. To find the velocity, we can use the concept of acceleration due to gravity.

When the volleyball is dropped from rest, it falls freely due to gravity. The acceleration due to gravity on Earth is approximately 9.8 m/s². By using the equation of motion:

v = u + at

where:
v = final velocity (which is the velocity just before hitting the ground)
u = initial velocity (which is 0 m/s as it is dropped from rest)
a = acceleration due to gravity (which is 9.8 m/s²)
t = time taken (which is 1.30 s)

Plugging in the values, we can calculate v:

v = 0 + (9.8 m/s²) × (1.30 s)
v = 12.74 m/s

Now that we have the mass (m = 0.330 kg) and the velocity (v = 12.74 m/s), we can calculate the momentum:

Momentum (p) = (0.330 kg) × (12.74 m/s)
Momentum (p) ≈ 4.20 kg·m/s

Therefore, the magnitude of the momentum just before the volleyball hits the ground is approximately 4.20 kg·m/s.