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 = 70.0° angle, with a speed v0A = 9.50 m /s. The horizontal separation between the clowns as they leave the cannons is d = 6.50 m. Find the launch speed v0B and the launch angle B (>45.0°) for clown B.

Explanation / Answer

Before collision both the projectile will be in the air for same time say t .
Now height covered by both of them is 1.05 m
So for first projetile
s = ut + 0.5at2
1.05 = 9.5Sin70*t - 0.5*9.81*t2
on solving t = 1.693 seconds
So horizontal distance covered by projectile 1
Distance = 9.5Cos70*1.693 = 5.5 m
So the horizontal distance covered by the projectile 2 = 6.5 -5.5 = 1 m
therefore
VOBCos(theta) = 1
now
1.05 = VOBSin(theta)*1.693 - 0.5*9.81*1.6932
VOBSin(theta) = 8.924
On squaring and adding above equation we get
VOB = 8.98 m/s
and theta = 83.6 degree

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