Question

A body is dropped from the top of a tower it aquires avelocity 20 ms -1on reaching the ground calculate the height of tower (take g = 10ms-2)​

Answer 1

$\large\underline{\bigstar \: \: {\sf SoluTion:-}}$

Initial velocity (u) = 0

Final velocity (v) = 20 m/s

g = 10 m/s²

Height of tower = h

$\boxed {\sf{v^{2}=u^{2}=2gs}}$

$\sf {\Rightarrow 20^{2}-0=2 \times 10\ h}\\\\\\\sf {\Rightarrow h=20\ m}$

∴ Height of tower = 20 m

Answer 2

Given ,

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 m/s

Acceleration (a) = 10 m/s²

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

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

Thus ,

$\mapsto \sf {(0)}^{2} - {(20)}^{2} = 2 \times 10 \times s \\ \\ \mapsto \sf 20s = 400 \\ \\ \mapsto \sf s = \frac{400}{20} \\ \\ \mapsto \sf s = 20 \: m$

$\therefore \sf \underline{The \: height \: of \: tower \: is \: 20 \: m}$