Question

Figure shows a projectile thrown with speed u = 20 m/s at an angle ${30}^0$ with horizontal from the top of a building 40 m high. Then the horizontal range of projectile is

• A. $20\sqrt{3}m$
• B. $40\sqrt{3}m$
• C. $40m$
• D. $20m$

Answer 1

The correct option is B $40\sqrt{3}m$

Given that,

Speed $u=20m/s$

Angle $\theta ={30}^0$

Height $h=40m$

Now, along y axis

$u_y=u\sin {30}^0$

$u_y=20\times \frac{1}{2}$

$u_y=10m/s$

Now, we know that

$a=-10m/s^2$

$s=-40m$

Now, from equation of motion

$s=u_{y}t-\frac{1}{2}g{t}^2$

$-40=10t-5t^2$

$5t^2-10t-40=0$

$t^2-2t-8=0$

$t^2-(4-2)t-8=0$

$t(t-4)+2(t-4)=0$

$(t+2)(t-4)$

Now, neglect of negative value

So, $t=4s$

Now, the range is

$R=u\cos {30}^0\times t$

$R=20\times \frac{\sqrt{3}}{2}\times 4$

$R=40\sqrt{3}m$

Hence, the range is $40\sqrt{3}m$