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3 years ago

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Physics

1 answer:

earnstyle [38]3 years ago

8 0

Hi there!

Recall that:

GPE = mgh

KE = 1/2mv²

At the TOP of an object's trajectory, the GPE is at a maximum. There is NO kinetic energy at this moment.

At the bottom of the fall, the KINETIC energy is at a maximum while the GPE is a minimum (assuming the ground to be the zero-line.)

We can write this out:

GPE = KE

mgh = 1/2mv²

There is an 'm' on both sides, so you can divide both sides by 'm':

(mgh)/m = 1/2mv²/m

gh = 1/2v²

Solve for v.

gh = 1/2v²

Multiply both sides by 2:

2gh = v²

Take the square root:

v = √2gh

We can use the equation:

v = √2gh

g = acceleration due to gravity (9.8 m/s²)

h = length of fall (3.67 m)

v = velocity (m/s)

Plug in the knowns:

v = √2(9.8)(3.67) = 8.48 m/s

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