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Beamsbox
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Projectile Motion - V0 is 5 times v @ max height?!?
Thanks prior, all...
A projectile's launch speed is 5 times its speed at maximum heigth. Find launch angle theta0.
Not sure which are relevant on this one, but the given for the section are:
x - x0 = (v0 cos theta0)t,
y - y0 = (v0 sin theta0)t - 1/2gt2,
vy = v0 sin theta0 - gt,
vy2 = (v0 sin theta0)2 - 2g(y - y0)
So, I've drawn the path, defining the maximum heigth as vm.
At vm, the vertical component, vmy, must be 0.
vmx must be constant and would equal the same as it started,
vmx = vox
In the problem, the initial velocity, v0 is 5 times vm, so
v0 = 5(vm) = 5(vox)
Not sure where to go from here, I tried to break vo into its components, but just ended up with voy2 = 24vox2
Any ideas?
Thanks again!
Thanks prior, all...
Homework Statement
A projectile's launch speed is 5 times its speed at maximum heigth. Find launch angle theta0.
Homework Equations
Not sure which are relevant on this one, but the given for the section are:
x - x0 = (v0 cos theta0)t,
y - y0 = (v0 sin theta0)t - 1/2gt2,
vy = v0 sin theta0 - gt,
vy2 = (v0 sin theta0)2 - 2g(y - y0)
The Attempt at a Solution
So, I've drawn the path, defining the maximum heigth as vm.
At vm, the vertical component, vmy, must be 0.
vmx must be constant and would equal the same as it started,
vmx = vox
In the problem, the initial velocity, v0 is 5 times vm, so
v0 = 5(vm) = 5(vox)
Not sure where to go from here, I tried to break vo into its components, but just ended up with voy2 = 24vox2
Any ideas?
Thanks again!