Physics

A pebble is thrown vertically upwards with a speed of 20 m s^-2. How high will it be after 2 s ? (Take g = 10 m s^-2)

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As we know, the equation of motion

h = ut - $\dfrac{1}{2}$ gt² (a = -g, as movement is against gravity)

Given,

t = 2 s

g = 10 m s^-2

u = 20 m s^-2

Substituting the values in the formula, we get,

$h = (20 \times 2) - (\dfrac{1}{2} \times 10 \times 2^2) \\[0.5em] h = 40 - (5 \times 4) \\[0.5em] h = 20 m \\[0.5em]$

Hence, greatest height reached by the ball = 20 m

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