Question

At $t=0$ a projectile is fired from a point $O$ (taken as origin) on the ground with a speed of $50m/$s at an angle of ${53}^{\circ }$ with the horizontal. It just passes two points $A$ and $B$ each at height $75m$ above horizontal as shown.

The horizontal separation between the points $A$ and $B$ is
(Take $g=10m/s^2$)

• A. $30m$
• B. $60m$
• C. $90m$
• D. $120m$

Answer 1

The correct option is B $60m$


$u_y=u\sin {53}^{\circ }=50\times \frac{4}{5}=40m/s$
By using third equation of motion
$v_y^2=u_y^2-2g\left(75\right)u_y^2={40}^2-2\left(10\right)\left(75\right)=1600-1500=v_y=\pm 10m/s$
During upward journey:
$v_y=u_y-g\left(t_1\right)10=40-g{t}_1\Rightarrow t_1=3s$
During downward journey;
$v_y=u_y-g{t}_2-10=40-g{t}_2\Rightarrow t_2=5s\Rightarrow \Delta t=5-3=2sAB=u_x\times \Delta t=u\cos \theta \Delta t=50\times \frac{3}{5}\times 2=60m$