# What is the minimum initial speed needed for a shot of 3187 ft?

In Sussex County, Delaware, a post-Halloween tradition is "Punkin Chunkin," in which contestants build cannons, catapults, trebuchets, and other devices to launch pumpkins and compete for the greatest distance. Though hard to believe, pumpkins have been projected a distance of 4086 feet in this contest

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- gp4rtsLv 71 decade agoFavorite Answer
The distance a projectile travels if launch at velocity v0 and angle θ is

d = v0²/g * sin2θ

The maximum d occurs at θ = 45º where sin2θ = 1 and is

d = v0²/g

For a distance of 3187 ft,

v0 = √[3187*32.17] = 320 ft/s

This is 218 mph!

- Anonymous1 decade ago
It depends on the initial angle of the pumpkin when it is released.

Minimal speed for the distance would be achieved at an angle of 45°.

- Anonymous1 decade ago
speed of light, minimum

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