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## Homework Statement

A projectile is fired at 30 km/h at the angle θ in x direction of an x-y plane. Ignoring the effects of air resistance, find the maximum

range and the value of θ that gives this range

Part A: write down the initial horizontal and vertical components

of the velocity

Part B: Write two equations to give the horizontal and vertical positions of the

projectile as functions of time in terms of the initial speed, and the initial

angle .

Part C: Use these equations to show that the time the projectile will remain

airborne is given by [itex]t = \frac{2v}{g}sin\theta[/itex] .

Part D: Hence show that the range is given by [itex]\frac{2v^2}{g}sin(\theta)cos(\theta)[/itex]

Part E: Show, by differentiating, that the maximum horizontal distance will occur if fired at 45° and is given by [itex]x_{max}=\frac{v^2}{g}[/itex]

## Homework Equations

[tex]v=u+at \\

s=ut+\frac{1}{2}gt^2 \\[/tex]

## The Attempt at a Solution

Part A:

Horizontal - [itex] vcos\theta[/itex]

Vertical - [itex]vsin\theta[/itex]

Part B:

Horizontal - [itex]vcos(\theta)t[/itex]

Vertical - [itex]vsin(\theta)t+\frac{1}{2}gt^2 [/itex]

Part C is where I am stuck. I understand it but when I done it before I did it a different way (i think).

When I did it before (last year) I used v=at so t=v/a so [itex]t=\frac{vsin\theta}{g}[/itex] but as that only gives it until the max height (halfway through) it needs to be times by two and hence [itex] t=\frac{2v}{g}sin\theta[/itex]

But the question says to use what I did in Part B and I am not sure how that is done.

Part D: that is straightforward, just substituting what is found as t in Part C into the horizontal bit of Part B.

Part E: This part has my head spinning at the moment.

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