Velocity is supposed to be 19.8 m/s and X is supposed to be 14m And theta is sup

Question

Velocity is supposed to be 19.8 m/s and X is supposed to be 14m And theta is supposed to be around 71 degrees

I just want to know how to get these answers. Thank you! :)

from Boomo, the human cannonball, allows himself to be shot with a muzzle velocity, V, inclined at an angle 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, , and x. a cannon 873 2 m 20 m

Explanation / Answer

At A, he is moving horizontally.
hence speed will be horizontal component of initial vlocity.

vA = Vcos@

now he swings to B. (that means KE at A get converts into gravitational PE as his vertical height is changing)

using energy conservation,

gain PE = loss in KE

m g deltaH = Ki -Kf

m g L = m vA^2 / 2 - 0

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

and vA = Vcos@ = 6.26 ...........(i)

vertical height of B is 20m from initial position and at B his KE= 0 .

using energy conservation, (PE + KE = constant )

energy at inital position = energy at B
( PE + KE)

0 + m V^2 / 2 = m g (20) + 0

V =sqrt(2 x 9.8 x 20) = 19.8 m/s .......Ans(INitial velocity)

and using equation(i)

vA = Vcos@ = 6.26

cos@ = 6.26/ 19.8

@ = 71.6 deg .......Ans

horizontal distance of A is half of range.

d = R/2 = [ v^2 sin(2@)/g] / 2

    = (19.8^2) (sin(2*71.6)) / (2 x 9.8) = 12 m

and x = d + 2 = 12 + 2 = 14 m .........Ans

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