Question

A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall? $(g=9.8m/s^2)$

Answer 1

Given :

Initial velocity u = +20 m/s

Find velocity v = 0 m/s

Acceleration due to gravity = g = a = 9.8 m/$s^2$

To find: Height reached by the stone = s =?

From 3rd equation of motion we get,

$v^2-u^2$ = 2as

$0^2-(+20{)}^2$= 2 × 9.8 × s

- 400 = 2 × 9.8 × s

s = −400/(2×9.8) = – 20.4 m

The negative sign shows that the direction of displacement and gravity acceleration is the opposite direction.

Thus, the height reached by the stone = 20.4 m.