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 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
$2.0 $ and $3.0 \,sec$
Only at $3.0 \,sec$
Only at $2.0 \,sec$
After $1$ and $2\, sec$
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
$h = ut - \frac{1}{2}g{t^2} \Rightarrow 96 = 80t - \frac{{32}}{2}{t^2}$
$ \Rightarrow {t^2} - 5t + 6 = 0 \Rightarrow t = 2\,sec$ or $ 3\, sec$
Other Language