- Home
- Standard 11
- Physics
At $t = 0$ a projectile is fired from a point $O$(taken as origin) on the ground with a speed of $50\,\, m/s$ at an angle of $53^o$ with the horizontal. It just passes two points $A \& B$ each at height $75 \,\,m$ above horizontal as shown. The horizontal separation between the points $A$ and $B$ is ........ $m$
$30$
$60$
$90$
None
Solution
$\theta=53^{\circ}$
$u=50 \mathrm{m} / \mathrm{s}$
$y=75 \mathrm{m}$
$y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}$
$75=x \tan 53^{\circ}-\frac{10 x^{2}}{2 \times 50 \times 50 \times \cos ^{2} 53^{\circ}}$
$75=\frac{4 x}{3}-\frac{10 x^{2} \times 25}{2 \times 2500 \times 9}$
$\Rightarrow \quad 75=\frac{4 x}{3}-\frac{x^{2}}{180}$
$75 \times 180=(4 \times 60) x-x^{2}$
$x^{2}-240 x+13500=0$
$x=\frac{240 \pm \sqrt{57600-4(1)(13500)}}{2(1)}$
$x=\frac{240 \pm \sqrt{57600-54000}}{2}$
$x=\frac{240 \pm \sqrt{3600}}{2}$
$x=\frac{240 \pm 60}{2}$
$x_{1}=150 \mathrm{m}$
$x_{2}=90 \mathrm{m}$
$\Delta x=60 \mathrm{m}$
