Question

At $t=0$ a projectile is fired from a point $\text{O}$ (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 the 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 $60m$


Here, $u_y=u\sin {53}^{\circ }=50\times \frac{4}{5}=40\text{m/s}$

By using third equation of motion $v_y^2=u_y^2+2a_{y}h$

Here $u_y=40\text{m/s},a_y=-g\text{and}h=75\text{m}$

$v_y^2=u_y^2-2g\left(75\right)$

$v_y^2={40}^2-2\left(10\right)\left(75\right)$

$v_y^2=1600-1500$

$\Rightarrow v_y=\pm 10m/s$

During upward journey, using

$v_y=u_y+a_{y}t$

$v_y=u_y-g\left(t_1\right)$

$10=40-g{t}_1$

$\Rightarrow t_1=3s$

During downward journey, using

$v_y=u_y+a_{y}t$

$v_y=u_y-g{t}_2$

$-10=40-g{t}_2$

$\Rightarrow t_2=5s$

$\Rightarrow \Delta t=5-3$

$\Rightarrow \Delta t=2s$

$AB=u_x\times \Delta t=u\cos \theta \Delta t=50\times \frac{3}{5}\times 2=60m$