Hobbyists build a compressed air powered cannon which is able to launch a pumpkin a horizontal distance of 2700 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, in m/s?

Question

Physics

Giovanni Hansen

8 September, 00:02

Hobbyists build a compressed air powered cannon which is able to launch a pumpkin a horizontal distance of 2700 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, in m/s?

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Answers (1)

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Jaylin Middleton · Answer 1

89.81 m/s

Explanation:

We are that the maximum horizontal distance (Range) the pumpkin can go is 2700ft (822.96 m).

Range is given as:

R = u²sin2θ/g

Where u is initial velocity, g is acceleration due to gravity, θ is angle of projection.

When range is maximum, that means the angle is 45 degrees. This means that:

sin2θ = 1

Range will now be:

R = u²/g

822.96 = u²/9.8

u² = 9.8 * 822.96

u² = 8065.008

u = 89.81 m/s