Question

A projectile takes off with an initial velocity of $20\text{m/s}$ at an angle of elevation of ${45}^{\circ }$. It is just able to clear two hurdles of height $4\text{m}$ each, separated from each other by a distance $d$. What is the value of $d$? (Take $g=10{\text{m/s}}^2$)

• A. $26\text{m}$
• B. $13\text{m}$
• C. $31\text{m}$
• D. $37\text{m}$

Answer 1

The correct option is C $31\text{m}$


Given that,
Initial velocity $\left(u\right)=20\text{m/s}$
$u_x=20\cos {45}^{\circ }=10\sqrt{2}\text{m/s}$.
$u_y=20\sin {45}^{\circ }=10\sqrt{2}\text{m/s}$.
Height of hurdle $\left(h\right)=4\text{m}$
The projectile will cross $4\text{m}$ hurdle twice as shown in the image. Now, time at which projectile crosses the hurdles.
$h=u_{y}t-\frac{1}{2}g{t}^2$
$\Rightarrow 4=10\sqrt{2}t-\frac{1}{2}\times 10t^2$
$\Rightarrow 4=10\sqrt{2}t-5t^2$
$\Rightarrow 5t^2-10\sqrt{2}t+4=0$
By solving the quadratic equation, we get
$t_1=\frac{\sqrt{30}+5\sqrt{2}}{5}$ and $t_2=\frac{-\sqrt{30}+5\sqrt{2}}{5}$
Now, distance $d$ between hurdles
$d=(t_1-t_2)u_x$
$\Rightarrow d=(\frac{\sqrt{30}+5\sqrt{2}}{5}-\left(\frac{-\sqrt{30}+5\sqrt{2}}{5}\right))\times 10\sqrt{2}$
$\Rightarrow d=\frac{2\sqrt{30}}{5}\times 10\sqrt{2}$
$\Rightarrow d=8\sqrt{15}=30.98\approx 31\text{m}$
Hence, option $\left(c\right)$ is the correct answer.