5.0 kg ball on a stirng is lifted to a height of 2.5m. It it swung down, what...

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...is the velocity at the bottom? Pendulum swinging downwards, what is the velocity at the bottom of the swing arc?
Please dont just tell me the answer, give me the process of how you got it as well.
 
This is a conservation of energy problem.

The start of swing potential energy PE = mgh = 1/2 mV^2 = KE the kinetic energy at the bottom of the swing. Thus V^2 = 2gh and V = sqrt(2gh); where g = 9.81 N/kg, and h = 2.5 m in your case. Note m = 5 kg has no bearing on the problem; meaning V will be the same for any given mass.

You can plug and chug now you know the physics.
 
mgh=1/2(mv^2)

so if it was lifted 2.5 m then thats your height.
so (5.0 kg)(9/8m/s^2)(2.5m) = 122.5 J
so at the top of the swing your kinetic energy is O and Your gravitional energy is 122.5 J
but when its at the bottom all converts turn kinetic energy so kinetic energy equals 122.5 J and gravitational equal 0

122.5 = 1/2((5.0kg)v^2)
so your velocity at the bottom of the pendulum is 7 m/s
 
mgh=1/2(mv^2)

so if it was lifted 2.5 m then thats your height.
so (5.0 kg)(9/8m/s^2)(2.5m) = 122.5 J
so at the top of the swing your kinetic energy is O and Your gravitional energy is 122.5 J
but when its at the bottom all converts turn kinetic energy so kinetic energy equals 122.5 J and gravitational equal 0

122.5 = 1/2((5.0kg)v^2)
so your velocity at the bottom of the pendulum is 7 m/s
 
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