Prove that the distance s metres, which a body falling from rest covers in time t seconds is 4.9 times the square of the time t.

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Solution

From second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
Putting $u = 0, a = 9.8$ ,
$s = 4.9 t^{2}$

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t

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4.8

times the square of the time

t

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When an object falls from a height, the distance s (metres), it travels in time t (seconds) is given by the equation s = 4.9t². Is the distance travelled proportional to time?

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Prove that the distance s metres, which a body falling from rest covers in time t seconds is 4.9 times the square of the time t.

From second equation of motion,s=ut+12at2Putting u=0, a=9.8,s=4.9t2

From second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
Putting $u = 0, a = 9.8$ ,
$s = 4.9 t^{2}$