Boomo, the human cannonball, allows himself to be shot with a muzzle velocity, V

Question

Boomo, the human cannonball, allows himself to be shot with a muzzle velocity, V, from a cannon inclined at an angle theta to the horizontal. When he is moving horizontally (point A), he grabs trapeze T, of length 2 meters, and swings up to point B on the platform, where the trapeze is horizontal. Point B is at a horizontal distance x and vertical distance 20 meters from the mouth of the cannon. In order to make the ace more dramatic, Boomo makes sure that the muzzle velocity is just great enough so that he just reaches point B and does not go above it. Find V, theta, and x.

Explanation / Answer

At point A, his velocity is vA then applying energy conservation from A to B to find vA.

m vA^2 /2 = m g L

vA = sqrt(2gL)= sqrt(2 x 9.8 x 2) = 6.26 m/s

at A, He has only horizontal velocity hence vx = vcos@ = 6.26 m/s

Now applying energy conservation from ground to A:

0 + m v^2 /2 = m g (20 - 2) + m vA^2 /2

v^2 = (2 x 9.8 x 18) + (6.26^2)

v = 19.8 m/s ......Ans (Initial Velocity )

and vA = vcos@

cos@ = 6.26 / 19.8 = 71.57 deg .......Ans

x is the half of same level range.

x = v^2 sin(2@) / 2g = (19.8^2 sin(2*71.57)) / (2 * 9.8) = 12 m ...........Ans

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