Physics

A body is dropped from the top of a tower. It acquires a velocity 20 m s^-1 on reaching the ground. Calculate the height of the tower. (Take g = 10 m s^-2)

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As we know, from the equation of motion,

v² = u² + 2gs

u = 0

v = 20 m s^-1

g = 10 m s^-2

Substituting the values in the formula, we get,

$20^2 = 0^2 + (2 \times 10 \times s) \\[0.5em] 400 = 20 s \\[0.5em] \Rightarrow s = \dfrac{400}{20} \\[0.5em] \Rightarrow s = 20 m \\[0.5em]$

Hence, the height of the tower = 20 m

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