Question

A ball is dropped from the edge of a roof how long does it take to fall 4.9 metre given acceleration is equal to 9.8 metre per second square

Answer 1

Given,
height of the roof= 4.9 metres
acceleration due to gravity=9.8 metres(g) we know,

$s = ut + \frac{1}{2} g {t}^{2}$
here initial velocity='0'[ since the ball is at rest]
so,substitute the values in given formula,
Here,
S=4.9
u=0
T=?

$4.9 = 0(t) + \frac{1}{2} \times 9.8 \times {t }^{2}$

$4.9 = 0 + 4.6 \times {t}^{2}$

$4.9 = 4.6( {t}^{2})$
${t}^{2} = \frac{4.9}{4.6}$
$t = \sqrt{1.06}$
therefore, T =1 SEC [Approximately]

Answer 2

Explanation:

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