A ball of mass .3 kg is dropped from a height of 10m. Its momentum when it strikes the ground is?

I know p=mv
how would I find velocity for this problem?

Thanks in advance for your help.

Use conservation of energy.

vf^2=Vi^2+2gh
vf= sqrt (2*9.8*10)

potential energy = kinetic energy

mgh=1/2 mv^2
cancel out mass
plug in gravity, height
solve for v

a ball of mass 5kg is dropped from a height 2meter.

P.e=k.e then,

Mgh=1\2mvsquare

To find the velocity of the ball when it strikes the ground, you can use the principle of conservation of energy. The potential energy of the ball at the top (when it is dropped) is converted to its kinetic energy when it hits the ground.

The potential energy (PE) of the ball can be calculated using the formula:

PE = m * g * h

Where:
m is the mass of the ball (0.3 kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height the ball was dropped from (10 m)

So, substituting these values into the formula:

PE = (0.3 kg) * (9.8 m/s^2) * (10 m)
PE = 29.4 Joules

Now, the kinetic energy (KE) of the ball can be calculated as:

KE = 1/2 * m * v^2

Where:
m is the mass of the ball (0.3 kg)
v is the velocity of the ball when it hits the ground (which we want to find)

Since the potential energy is converted entirely to kinetic energy, we can equate the two:

PE = KE

29.4 Joules = 1/2 * (0.3 kg) * v^2

Rearranging this equation to solve for v:

v^2 = (2 * 29.4 Joules) / (0.3 kg)
v^2 = 196 m^2/s^2

Taking the square root of both sides to solve for v:

v = sqrt(196 m^2/s^2)
v = 14 m/s

So, the velocity of the ball when it strikes the ground is 14 m/s.