{ HARDER }

Full solutions (either when someone gets it, or next week):


In the diagram, a large number of projectiles are fired simultaneously from O, each with the same velocity V m/s, but different angles of projection theta, at a wall d metres from O. The projectiles are fired so they all lie in the same vertical plane perpendicular to the wall.

You may assume that the equations of motion at time t are given by:
[tex]x = Vtcos\theta[/tex] and [tex]y = -\frac{1}{2}gt^{2} + Vtsin\theta[/tex].

i) Using these two equations of motion, prove the relationship between the height y and time t is:

[tex]4y^{2} + 4gt^{2}y + (g^{2}t^{4} + 4x^{2} - 4v^{2}t^{2}) = 0[/tex]

ii) Show that the first impact at the wall occurs at time [tex]t = \frac{d}{V}[/tex] and that this projectile was fired horizontally.

iii) Hence, find where this projectile hits the wall.

iv) Show that for [tex]t \ \textgreater \ \frac{d}{V}[/tex], there are two impacts at time t, and that the distance between these is:

[tex]2\sqrt{V^{2}t^{2} - d^{2}}[/tex].

v) Given that V = 10 m/s and d = 10 metres, what are the initial angles of projection of the two projectiles that will strike the wall simultaneously [tex]20\sqrt{3}[/tex] metres apart.