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

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Solution

Given :

Initial velocity u = +20 m/s

Find velocity v = 0 m/sec

Gravity acceleration g = 9.8 m/ $s^{2}$

Distance s = ?

From 3rd equation of motion

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

$0^{2} - (+ 20)^{2}$ = 2 × 9.8 × s

- 400 = 2 × 9.8 × s

s = −400/2×9.8 = – 20.4 m

Negative sign shows direction of displacement and gravity acceleration are opposite direction.

the distance = 20.4 m

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