Question

A particle is projected at an angle Alpha to the horizontal which grazes with top of the Wall of height B located at a horizontal distance of A from point of projection. then H max of the projectile is?????

Answer 1

If v_0 is the initial velocity, then the maximum height of a flight can be calculated as

$$H = \frac{v_{0y}^2}{2g} = \frac{v_0^2 (\sin{\alpha})^2}{2g}$$

The laws of motion of a projectile along the x- and y-directions can be written as:

$$x(t) = v_0 \cos{\alpha} \cdot t,\\ y(t) = v_0 \sin{\alpha} \cdot t - \frac{g t^2}{2}$$

Substituting the A and B values (horizontal distance and height) and excluding the time, we obtain:

$$t = \frac{A}{v_0 \cos{\alpha}}, \\ B = v_0 \sin{\alpha} \cdot \frac{A}{v_0 \cos{\alpha}} - \frac{g}{2} \left( \frac{A}{v_0 \cos{\alpha}} \right)^2 = A \tan{\alpha} - \frac{g}{2} \left( \frac{A}{v_0 \cos{\alpha}} \right)^2$$

Solving this equation in respect to the initial velocity squared, we derive:

$$v_0^2 = \frac{g A^2}{2 (\cos{\alpha})^2 (A \tan{\alpha} -B)}$$

Substituting this value into the initial expession, we conclude:

$$H = \frac{ (A \tan{\alpha})^2}{4 (A \tan{\alpha} -B)}$$