Question

{ 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:

`x = Vtcos\theta` and
`y = -(1)/(2)gt^(2) + Vtsin\theta`.

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


`4y^(2) + 4gt^(2)y + (g^(2)t^(4) + 4x^(2) - 4v^(2)t^(2)) = 0`

ii) Show that the first impact at the wall occurs at time
`t = (d)/(V)` and that this projectile was fired horizontally.

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

iv) Show that for
`t \ \textgreater \ (d)/(V)`, there are two impacts at time t, and that the distance between these is:


`2\sqrt{V^(2)t^(2) - d^(2)}`.

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
`20√(3)` metres apart.

Answer 1

(i):

`x = Vtcos\theta`

`Vcos\theta = (x)/(t)`


`y = -(1)/(2)gt^(2) + Vtsin\theta`

`Vsin\theta = (y + (1)/(2)gt^(2))/(t)`


`V^(2)cos^(2)\theta + V^(2)sin^(2)\theta = (x^(2))/(t^(2)) + (y^(2) + gt^(2)y + (1)/(4)g^(2)t^(4))/(t^(2))`

`V^(2) = (x^(2))/(t^(2)) + (y^(2) + gt^(2)y + (1)/(4)g^(2)t^(4))/(t^(2))`

`t^(2)V^(2) = x^(2) + y^(2) + gt^(2)y + (1)/(4)g^(2)t^(4)`

`4t^(2)V^(2) = 4x^(2) + 4y^(2) + 4gt^(2)y + g^(2)t^(4)`

`4x^(2) + 4y^(2) + 4gt^(2)y + g^(2)t^(4) - 4t^(2)V^(2) = 0`

`4y^(2) + 4gt^(2)y + (gt^(2)t^(4) + 4x^(2) - 4t^(2)V^(2)) = 0`

(ii): Impact is when x = d.

`\text{Impact: } d = Vtcos\theta`

`t = (d)/(Vcos\theta)`

First impact occurs when t is minimised.
This means that Vcos theta is maximised, which means cos theta = 1, and theta = 0

`\therefore \text{First impact occurs at } \theta = 0\text{: }t = (d)/(V(1)) = (d)/(V)`

i'll do the rest later.