Time, Speed & Distance
Aptitude

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Q.

A ball is projected vertically up-wards reaches, at the end of '$t$' seconds, an elevation of '$s$' feet where $s=160t−16t^2$. The highest elevation is

 A.

800

 B.

640

 C.

400

 D.

320

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Solution:
Option(C) is correct

Given that:

$s=160t−16t^2.$

Highest value of $S$ will be reached when \(\dfrac{dS}{dt}=0\)

$⇒ 160−32t=0 $

$⇒ t=5$ sec

Thus $S=(160\times5)−(16\times 5\times 5) $

$=800−400$= 400


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