what is the minimum initial speed of the pumpkin (just as it leaves the cannon) that is needed for it to reach this distance?
Hobbyists build a compressed air powered cannon which is able to launch a pumpkin a horizontal distance of 3600 ft. Assuming no air resistance, and assuming the pumpkin is launched at ground level, what is the minimum initial speed of the pumpkin (just as it leaves the cannon) that is needed for it to reach this distance?
- Old Science GuyLv 75 years agoFavorite Answer
when a projectile comes back to the same height as that from which it was fired
this is the range equation
R = (Vo^2 / g) sin 2 θ
without using too much complicated math
we can see that for any given Vo the max R will be when sin 2 θ is also max
and the max value of sin is 1
so sin 2 θ = 1 gives 2 θ = 90 and θ = 45
now we can go back and find the Vo at 45°, the optimal angle
3600 = Vo^2 / 32.2 (1)
Vo^2 = 115920
Vo = 340.5 ft/s
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