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FaraDazed
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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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