2.Motion in Straight Line
medium

A particle is projected up with an initial velocity of $80\;ft/\sec $. The ball will be at a height of $96\;ft$ from the ground after

A

$2.0 $ and $3.0 \,sec$

B

Only at $3.0 \,sec$

C

Only at $2.0 \,sec$

D

After $1$ and $2\, sec$

Solution

(a) $h = ut – \frac{1}{2}g{t^2} \Rightarrow 96 = 80t – \frac{{32}}{2}{t^2}$

$ \Rightarrow {t^2} – 5t + 6 = 0 \Rightarrow t = 2\,sec$ or $ 3\, sec$

Standard 11
Physics

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