Two cannons are mounted as shown in the drawing and rigged to fire simultaneousl

Question

Two cannons are mounted as shown in the drawing and rigged to fire simultaneously. They are used in a circus act in which two clowns serve as human cannonballs. The clowns are fired toward each other and collide at a height of 1.05 m above the muzzles of the cannons. Clown A is launched at A = 80.0° angle, with a speed v0A = 9.50 m /s. The horizontal separation between the clowns as they leave the cannons is d = 5.80 m. Find the launch speed v0B and the launch angle B (>45.0°) for clown B. Ask for the magnitude (? m/s) and direction( ?degree)

Explanation / Answer

The time it takes for the collision to occur is the time it takes either clown to reach an altitude of 1.05 m Since you are given data for clown A,

yA = v0Ay*t - 0.5*g*t²

v0Ay = v0A*sinA so

yA = v0A*sinA*t - 0.5*g*t²

4.9*t² - 9.5*sin80º*t + 1.05 = 0

4.9*t² - 9.36*t + 1.05 = 0

t = 1.79 s

The vertical trajectory for B is

yB = v0by*t - 0.5*g*t²

v0By = v0B*sinB

yB =v0B*sinB*t - 0.5*g*t²

1.05 = v0B*sinB*2.02 - 15.7

v0B*sinB = 16.75/1.79=9.35

In the horizontal direction

v0Ax*t + v0Bx*t = d

v0A*cosA + v0B*cosB = d/t

9.5*cos80º + v0B*cosB = 5.8/1.79

1.65 + v0B*cosB = 3.18 =3.24

v0B*cosB = 1.59 also from above
v0B*sinB = 9.35

Divide the two eqs

tanB = 5.88 = 24.40º

v0B = 1.59/(cos24.40º) = 0.037 m/s

there may be calculation mistake but concept is absolutely right

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