Question

The projectile is thrown at $25\sqrt{2}m/s$ at an angle of ${45}^{\circ }$, if the objective is just to clear both the posts, each with a height of $30m$, find the distance between the posts?

• A. 25 m
• B. 30 m
• C. 50 m
• D. 20 m

Answer 1

The correct option is A

25 m

Let's analyze the motion

Suppose the first post is $x$ units away from origin. Which indirectly implies the second post is at $(R-x)$ units away from origin (Refer the diagram)

$\overrightarrow{v_x}=vcos{45}^{\circ }=25\sqrt{2}\times \left(\frac{1}{\sqrt{2}}\right)=25\frac{m}{s}\uparrow$

$\overrightarrow{v_y}$ = $25j$

Motion along y-axis

$a$ = $-g\frac{m}{s^2}$

$v_y$ = $25\frac{m}{s}$

$h$ = $30m$

$S$ = $ut+\frac{1}{2}a{t}^2$

$30$ = $25t$ - $5t^2$

$5t^2-25t+30=0$

$t^2-5t+6=0$

$t^2-3t-2t+6=0$

$t(t-3)-2(t-3)=0$

$(t-3)(t-2)=0$

$t=2$ & $3$(what's the meaning of this?)

$2{}^{\prime }{t}^{\prime }s$ because there will be two times when the ball will be $30m$ above the ground in its projectile motion.

Motion along x-axis

$u$ = $25\frac{m}{s}$

$a$ = $0$

$s$ = $ut+\frac{1}{2}a{t}^2$

$s$ = $25t$

at $2s$

$x_1$ = $50m$

at $3s$

$x_2$ = $75m$

$x_2-x_1$ = $25m$

Required

Alternate Solution

Equation of trajectory:- $y$ = $xtan\theta -x^2\frac{tan\theta }{R}$ ; $y$ = $30$; $R$ = Range = $u^{2}sin\left(\frac{2\theta }{g}\right)$; $\theta$ = ${45}^{\circ }$ $u$ = $25\sqrt{2}$

Substitute:-

$x^2-125x+3750=0$

$\Rightarrow$ sum of roots $x_1+x_2=125$; product of roots $x_1{x}_2=3750$

we need $x_2-x_1$ = $\sqrt{(x_1+x_2{)}^2-4x_1{x}_2}$ $\Rightarrow 25m$