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A projectile is fired in such a way that its horizontal range is equal to 2 times its maximum height. What is?
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A projectile is fired in such a way that its horizontal range is equal to 2 times its maximum height. What is the angle of projection?

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A projectile is fired in such a way that its horizontal range is equal to 2 times its maximum height. What is?
A year ago
R = 2 H(max)
u^2 sin 2p / g = 2 [u^2 sin^2 p] / 2g
sin 2p = sin^2 p
2 sin p cos p -  sin^2 p = 0
sin p [2 cos p - sin p] = 0
sin p = can not be zero as it makes p = 0
--------------------------------
2 cos p = sin p
tan p = 2
p = 63.43 degree above horizontal


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A projectile is fired in such a way that its horizontal range is equal to 2 times its maximum height. What is?
A year ago
This question is trickier than it looks. But i think i might have the answer. But then again I could be wrong.
Firstly lets denote the angle as x for convenience.
The projecticle is initially travelling at V m/s at an angle of x degrees. The acceleration acting on this projectile is - g
The horizontal distance at any given time can be formulated as:
x = Vtcos(x)
The vertical distance at any given time can be formulated as:
y = Vtsin(x) - gt²/2
Both of these formulas are derived from s = ut + at²
In order to work out the range, we must remember that when the projectile has achieved its full range, the vertical component of distance (ie the height) would be 0. So we must substitute y =0 into the y equation and calculate the times when y =0
Therefore:
0 = Vtsin(x) - gt²/2 = t[Vsin(x) - gt/2]
We have two solutions. t = 0 or Vsin(x) - gt/2 = 0. We take the second solution as the first solution simply refers to the intial position of the projectile.
Vsin(x) - gt/2 = 0  therefore: 2Vsin(x)/g = t
We now know that the range of the projectile is achieved when t = 2Vsin(x)/g
Substitute this into the x equation to obtain
x = 2V²sin(x)cos(x)/g
This equation can be used to calculate the range of the projectile.
We must now find an equation for the maximum height of the projectile.
When maximum height occurs, the vertical component of velocity = 0
Using the equation v = u + at we get:
0 = Vsin(x) - gt                
Therefore: t = Vsin(x)/g
We now know the time when maximum height occurs. Now we must substitute this into the y equation:
y(max) = V²sin²(x)/g - (g/2)*(Vsin(x)/g)²
y(max) = V²sin²(x)/g - (g/2)*(V²sin²(x)/g²)
y(max) = V²sin²(x)/g - V²sin²(x)/2g                        - we can use simple algebra to make this one fraction
y(max) = V²sin²(x)/2g

We now have formulas for the range and the maximum height. The stipulation given to us is that the horizontal range is two times the maximum height. Ie x = 2y(max)
Therefore:
y(max) = V²sin²(x)/2g  Therefore: 2y(max) = V²sin²(x)/g
Now x = 2y(max)
Therefore:
2V²sin(x)cos(x)/g = V²sin²(x)/g
As we are dividing both sides by g they both effectively cancel each other out. So..
2V²sin(x)cos(x) = V²sin²(x)  
Dividing throughout by V²sin(x) gives:
2cos(x)=sin(x)
Divide throughout by cos(x) gives:
2 = sin(x)/cos(x)
Remember sin(x)/cos(x) = tan(x)
Therefore:
2 = tan(x)
x = tan^-1(2)
x = 63.43494882
Therefore when the horizontal range is equation to 2 times the maximum height the angle of projection must be 63.43494882

Sorry for the ridiculously long answer, but thats how much working out I had to do to obtain my answer.


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