Two cannons are mounted as shown in the drawing and rigged tofire simultaneously

Question

Two cannons are mounted as shown in the drawing and rigged tofire simultaneously. They are used in a circus act in which twoclowns serve as human cannon balls. The clowns are firedtoward each other and collide at a height of 1.00m above themuzzles of the cannons. Clown A is launched at a75.0o angle, with a speed of 9.00m/s. Thehorizontal separation between the clowns as they leave the cannonsis 6.00m. Find the launch speed v0B and the launchangle B (>45.0o) for clown B. answer: 8.79m/s, 81.5o

Explanation / Answer

The idea is that you have to find (many small steps): .    time for the first clown to reach height of 1.00 m .    horizontal distance traveled by first clown .    then horiz dist traveled by second clown .    initial horizontal velocity of second clown . initial vertical velocity of second clown .    total velocity of second clown (magnitude andangle) . So... . use vertical equation of motion for first clown to findtime: .          1.00 = 9.00 sin75 * t   - (1/2) * 9.80 *t2     quadratic equation .                4.9 t2   - 8.69334 t   + 1.00 = 0 .      t =   1.6505 seconds . Then use horizontal equation to get horiz distance: .     d = 9.00 cos75 * 1.6505 =   3.8446 meters . So the second clown must travelhorizontally:    6.00 - 3.8446=    2.1554 meters . And his horizontal speed must be:   distance /time = 2.1554 / 1.6505 = 1.3059 m/s . His vertical speed can be found from the vertical motionequation: .      1.00 = vertical speed *1.6505 - (1/2) * 9.80 * 1.65052 .     vertical speed =   8.6933 m/s . Magnitude of initial velocity of second clown = (1.30592 + 8.69332)1/2   =     8.79 m/s .     angle      arctan (8.6933 / 1.3059) =     81.5 degrees .     vertical speed =   8.6933 m/s . Magnitude of initial velocity of second clown = (1.30592 + 8.69332)1/2   =     8.79 m/s .     angle      arctan (8.6933 / 1.3059) =     81.5 degrees

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