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 0.95 m above the muzzles of the cannons. Clown A is launched at ?_A = 76.0

Answer 1

Clown A is launched 75 degrees above the horizontal with a speed of 9 m/s and meets Clown B 1 m above the cannon.

x = xo + vo*t + 1/2 a t^2,

where x is the final position,

xo is the initial position,

vo is the initial velocity,

t is the change in time,

and a is the acceleration.

1.0 m = 0 m + (9sin 75 m/s)t + 1/2 (-9.8 m/s^2)t^2

1 = 8.693t - 4.9t^2

4.9t^2 - 8.693t + 1 = 0

Using the quadratic equation:

t = [8.693 +/- sqrt( 8.693^2 - 4(4.9)(1)) ] / 2(4.9)

t = .124 s, 1.651 s

The later time refers to what would happen if the clown didn't stop. He would fall back to the ground, and at some point be 1m high up again. So the collision occurs after .124 seconds.

Horizontally, the clown would be:

x = 0 m + (9cos 75 m/s)(.124 s) + 1/2(0 m/s^2)(.124 s)^2

x = .288 m from A's cannon.

For Clown B:

Vertically:

1 m = 0 m + (v sin B m/s)(.124 s) + 1/2 (-9.8 m/s^2)(.124 s)^2

1 = .124 v sin B - .075

1.075 = .124 v sin B

8.693 = v sin B

Horizontally:

6 m - .288 m = 0 m + (v cos B m/s)(.124 s) + 1/2 (0 m/s^2)(.124 s)^2

5.712 = .124 v cos B

46.195 = v cos B

a)

(v cos B)^2 + (v sin B)^2 =

v^2 (cos^2 B + sin^2 B) =

v^2

v^2 = 8.693^2 + 46.195^2

v = 47 m/s

b)

8.693 = v sin B

8.693 = 47 sin B

sin B = .185

B = 10.7 degrees