A bullet is fired from the bottom of the inclined plane at angle $θ = 37^{o}$ 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 $θ$ at which the range will be maximum; (e) maximum range.

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Solution

a.)Taking axis system as shown in fig.At highest point, $v_{y} = 0.$
Therefore ,
$v_{y}^{2} = u_{y}^{2} + 2 a_{y} y$
$\Rightarrow 0 = (30)^{2} - 2 g cos 30^{o} y$
$y = 30 \sqrt{3} m$ (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.) $θ = \frac{π}{4} - \frac{30^{o}}{2} = 45^{o} - 15^{o} = 30^{o}$
e.) $\frac{u^{2}}{g (1 + sin β)} = \frac{50 \times 50}{10 (1 + \frac{1}{2})} = \frac{2500}{15} = \frac{500}{3} m$

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