# 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?

### 1 Answer

- 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

round appropriately

When you get a good response,

please consider giving a best answer.

This is the only reward we get.

from where did you got that 32.2 in place of g