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If a coin is dropped from a height of 80 m, how fast will it be travelling when it hits the ground? What assumptions, if any, do you need to make?

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1 Answer

Sep 13, 2017

$v_{f} =$ $v_f=$ - $40 \frac{m}{s} ˆ j$ $40m/s hatj$

Explanation:

In case of free fall

$v_{f} = \sqrt{2 \cdot g \cdot h}$ $v_f=sqrt(2*g*h)$

so $v_{f} = \sqrt{2 \cdot 10 \cdot 80}$ $v_f=sqrt(2*10*80)$
$v_{f} = 40 \frac{m}{s}$ $v_f=40m/s$ .

we didn't assumed anything .

The derivation of the equation is from 3rd equation of motion.

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