Question

{ HARDER }

Answer 1

(i):
$x = Vtcos\theta$
$Vcos\theta = \frac{x}{t}$

$y = -\frac{1}{2}gt^{2} + Vtsin\theta$
$Vsin\theta = \frac{y + \frac{1}{2}gt^{2}}{t}$

$V^{2}cos^{2}\theta + V^{2}sin^{2}\theta = \frac{x^{2}}{t^{2}} + \frac{y^{2} + gt^{2}y + \frac{1}{4}g^{2}t^{4}}{t^{2}}$
$V^{2} = \frac{x^{2}}{t^{2}} + \frac{y^{2} + gt^{2}y + \frac{1}{4}g^{2}t^{4}}{t^{2}}$
$t^{2}V^{2} = x^{2} + y^{2} + gt^{2}y + \frac{1}{4}g^{2}t^{4}$
$4t^{2}V^{2} = 4x^{2} + 4y^{2} + 4gt^{2}y + g^{2}t^{4}$
$4x^{2} + 4y^{2} + 4gt^{2}y + g^{2}t^{4} - 4t^{2}V^{2} = 0$
$4y^{2} + 4gt^{2}y + (gt^{2}t^{4} + 4x^{2} - 4t^{2}V^{2}) = 0$

(ii): Impact is when x = d.
$\text{Impact: } d = Vtcos\theta$
$t = \frac{d}{Vcos\theta}$

First impact occurs when t is minimised.
This means that Vcos theta is maximised, which means cos theta = 1, and theta = 0
$\therefore \text{First impact occurs at } \theta = 0\text{: }t = \frac{d}{V(1)} = \frac{d}{V}$

i'll do the rest later.