Question

A bullet is fired from the bottom of the inclined plane at angle $\theta ={37}^o$ with the inclined plane. The angle of incline is ${30}^o$ with the horizontal. Find the (a) position of the maximum height of the bullet from the inclined plane; (b) time of light; (c) range along the incline; (d) the value of $\theta$ at which the range will be maximum; (e) maximum range.

Answer 1

a.)Taking axis system as shown in fig.At highest point,$v_y=0.$

Therefore ,

$v_y^2=u_y^2+2a_{y}y$

$\Rightarrow 0=(30{)}^2-2g\cos {30}^{o}y$

$y=30\sqrt{3}m$ (maximum height)....................(1)

b.) Again for $x$ co-ordinate,

$v_y=u_y+a_{y}t$

$0=30-g\cos {30}^o\times t\Rightarrow t=2\sqrt{3}s$

Time of flight $T=2\times 2\sqrt{3}s$

c.) range ,$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$=40\times 4\sqrt{3}-\frac{1}{2}g\sin {30}^o\times (4\sqrt{3}{)}^2$

$40(4\sqrt{3}-3)m$

d.)$\theta =\frac{\pi }{4}-\frac{{30}^o}{2}={45}^o-{15}^o={30}^o$

e.)$\frac{u^2}{g(1+\sin \beta )}=\frac{50\times 50}{10(1+\frac{1}{2})}=\frac{2500}{15}=\frac{500}{3}m$