Question

A projectile shot at an angle of 60° above the horizontal strikes a building 25m away at a point 16m above the point of projection. Find the magnitude and direction of the velocity of the projectile as it strikes the building

Answer 1

Let the x axis be directed from the point of the start of the motion towards the base of the building and the y axis be perpendicular to the x axis and directed vertically.

The projection of the initial velocity onto x axis is $v_x = v_0\cos\alpha$ and onto y axis is $v_y=v_0\sin\alpha - gt.$ Therefore, x coordinate will be $x=v_0\cos\alpha t, \; y = v_0\sin\alpha t - \dfrac{gt^2}{2}.$

We may write the system

$\begin{cases} x=v_0\cos\alpha t, \\ y = v_0\sin\alpha t - \dfrac{gt^2}{2}. \end{cases} \;\;\; \begin{cases} 25=v_0\cos60^\circ t, \\ 16 = v_0\sin60^\circ t - \dfrac{9.8t^2}{2}. \end{cases} \;\;\; \begin{cases} t=\dfrac{25}{v_0\cos60^\circ}, \\ 16 = 25\tan60^\circ - \dfrac{9.8\cdot\dfrac{25^2}{v_0^2\cos^260^\circ}}{2}. \end{cases}$

Therefore, $v_0 = 21.2\,\mathrm{m/s}, \; t= 2.36\,\mathrm{s}.$ The velocity components will be

$v_x = v_0\cos\alpha = 10.6\,\mathrm{m/s}, \; v_y = v_0\sin\alpha - gt = -4.77\,\mathrm{m/s}.$ So the total velocity will be $\sqrt{v_x^2+v_y^2} = 11.6\,\mathrm{m/s}$ . The angle between the direction of x axis and the velocity is $\tan\beta = \dfrac{v_y}{v_x} = -0.45, \beta = -24^\circ.$