Question

A vertical wall of height 3.75 m is at a horizontal distance of 5 m from a point O on the ground. A particle is projected from O with initial speed 10 m/s at an angle of elevation x. Given that the particle just clears the wall, find
A) the value of tanx
B) the speed of the particle as it passes over the wall
C) the horizontal distance of the particle from the wall when it again reaches the level of O. Take g as 10 m s^-2

Answer 1

A). since,

$$Y=\tan(x)X+\frac{gX^2}{2u^2}(1+\tan^2(x))$$

Now,

$$3.75=5\tan(x)+\frac{5}{4}(1+\tan^2(x))$$

Let,$p=\tan(x)$ ,thus

$$p^2+p-2=0\implies (p+2)(p-1)=0\\\implies p=1,-2$$

By convention, since $x$ is positive and $0^\circ<x\leq 90^\circ$ ,hence $p=1\iff \tan(x)=1$

B).As $5=u\cos(x)t\implies t=\frac{1}{\sqrt{2}}$ and $v_y=u\sin(x)-gt\implies v_y=0$ ,thus

$$v=\sqrt{v_x^2+v_y^2}=5\sqrt{2}m/s$$

C).When the particle reaches the ground

$$3.75=\frac{1}{2}gt^2\implies t=\frac{\sqrt{3}}{2}$$

And,

$$x_1=u\cos(x)t=5\sqrt{\frac{3}{2}}\approx6.12m$$