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A ball released from a certain height falls 5m in 1 second; in 4 seconds, it falls through...? Take $g=10m/s^2$

Answer : 80m

My attempt: Since the body is released from a certain height, $u=0$

Therefore, $H=(0.5)gt^2$

In 4 seconds, displacement of the particle from the point of release = $(0.5)g(4)^2$ =80m

Now, I don't understand what is the requirement of the first information that the ball falls 5metres in one second. Am I missing something or the information is not required at all?

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2 Answers 2

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The problem probably is illustratory. It's trying to show how constant acceleration motion gives you $t^2$ variation for the displacement. The displacement undergone in $4$ seconds is $16$ times the displacement undergone in $1$ second. (Beginners' intuition would suggest $4$ times.)

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  • $\begingroup$ Random downvotes on this website make me laugh so much. $\endgroup$
    – Hritik Narayan
    Commented Jun 2, 2017 at 11:41
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It does seem there is superfluous information here. Just knowing the ball falls 5 meters in one second is enough to compute the drop height after 4 seconds.

It's possible the key here is that the extra piece of information tells us that air drag can be neglected (otherwise the ball would have fallen less than 5 m in a second, given g=10m/s$^2$).

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