Question

A particle is projected from a point on the horizontal plane so as to just clear two walls each of height $20m$ at distance $30m$ and $170m$ respectively from the point of projection. If $\alpha$ is the angle projection then

• A. $40\tan \alpha =51$
• B. $40\cot \alpha =51$
• C. $30tan\alpha =23$
• D. $30\cot \alpha =23$

Answer 1

Answer: (B)

Let the angle of projection be $\alpha$ .
Then
$30=u\cos \alpha t$
$\Rightarrow$ $t=\frac{30}{u\cos \alpha }t$ ..(i)
$\Rightarrow$ $2\theta =(u\sin \alpha )t-\frac{1}{2}g{t}^2$
$\Rightarrow$ $20=u\sin \alpha \left(\frac{30}{u\cos \alpha }\right)-\frac{1}{2}g{\left(\frac{30}{u\cos \alpha }\right)}^2$
$\Rightarrow$ $20=30\tan \alpha -\frac{900}{2}.\frac{g}{u^2{\cos }^2\alpha }$ ..(ii)
and $R=\frac{u^2\sin 2\alpha }{g}=200$
$\Rightarrow$ $u^2=\frac{200g}{\sin 2\alpha }$
On putting this value in Eq. (ii), we get $20=30\tan \alpha =\frac{900g\sin 2\alpha }{2\times 200g{\cos }^2\alpha }$
$\Rightarrow$ $20=30\tan \alpha -\frac{9\times 2\sin \alpha \cos \alpha }{4{\cos }^2\alpha }$
$\Rightarrow$ $20=30\tan \alpha -\frac{9}{2}\tan \alpha$
$\Rightarrow$ $20=\frac{(60-9)}{2}\tan \alpha$
$\Rightarrow$ $\frac{40}{\tan \alpha }=51$
$\Rightarrow$ $40\cot \alpha =51$