Question

An object is dropped from a height of 50 meters starting at rest. Using the principle of conservation of energy, What is its final velocity just before it hits the ground? What is the time it takes to fall?

Answer #1

Here we have given that,

Height H = 50 m

a) So that using the principle of energy conservation here

Potential energy at height H = kinetic energy just above the ground

MgH = 0.5 MV²

So that ,

Final velocity V will be here,

V = (2gH)⁰'⁵ = 31.3129 m/s

Hence the final velocity will be 31.3129 m/s

Now for the next part

b)

Now at the height of 50 m its initial velocity u is zero so that using law of kinematics her we have,

S = 0.5gt²

t = (50×2/9.8)^0.5 = 3.19438 sec

Hence time of fall will be 3.19438 sec

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