Question

od-ow
ground is:
An object is dropped from a height of 100 m.
Its velocity at the moment it touches the?
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Answer 1

Explanation:

when the body touches the ground...it has44.2m/s

Answer 2

Given ,

Displacement (s) = 100 m

Initial velocity (u) = 0 m/s

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

We know that , the Newton's third equation of motion is given by

$\large \sf \fbox{ {(u)}^{2} - {(v)}^{2} = 2as}$

Thus ,

$\sf \mapsto {(v)}^{2} = 2 \times 9.8 \times 100 \\ \\ \sf \mapsto v = \sqrt{19.6 \times 100} \\ \\ \sf \mapsto v = \sqrt{1960} \\ \\ \sf \mapsto v = 44.2 \: \: m/s$

$\therefore \sf \underline{The \: final \: velocity \: of \: object \: is \: 44.2 \: m/s}$