Question

If $4$ seconds be the time in which a projectile reaches a point $P$ of its path and $5$ seconds the time from $P$ till it reaches the horizontal plane through the point of projection. The height of $P$ above the horizontal plane will be [$g=9.8m/se{c}^2]$

• A. $98m$
• B. $49m$
• C. $196m$
• D. $147m$

Answer 1

The correct option is A $98m$

Time of flight$=4+5=9$ s
so,$\frac{2usin\theta }{g}=9$
or $usin\theta =\frac{9g}{2}$
if $h$ is the height of $P$, then
$h=\left(usin\theta \right)4-\frac{1}{2}g(4{)}^2$
$=\frac{9g}{2}\times 4-\frac{1}{2}g\times 16$
$=10g=98m$
hence,option $A$is correct.