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 1.00 m above the muzzles of the cannons. Clown A is launched at a 75.0 degree angle, with a speed of 9.00m/s. The horizontal separation between the clowns as they leave the cannons is 6.00 m. Find the launch speed v_0.8 and the launch angle theta_B(>45 degree) for clown B.

Explanation / Answer

clown A's vertical speed is 9*sin75 = 8.69 m/s
horizontal is 9*cos75 = 2.33 m/s
his y distance is +1.00 m
y = v(y)*t + 1/2*g*t^2
1 = 8.69t - 4.9t^2 (quadratic equation)
solving for t ( 1.65 s, 0.124 s )
the shorter time is on the way up - the horizontal distance traveled is so short by then, about 03 m, that clown B would have to be launched at such a bone-crunchingly high speed to make up the rest of the 6 m that I reject this time and make the reasonable assumption that both clowns meet on the downward parts of their trajectories.

so
x = 2.33 * 1.65 for clown A or 3.84 m
for clown B it is 6.00-3.84 = 2.16 m
since they are fired simultaneously and meet, time is the same for both
so
x = V*1.65 for B
2.16 = V*1.65
v(x) = 1.31 m/s
now
since they have the same y displacement in the same time
the upward velocities must be equal (you can solve the equation, but you'll get the same value)
v(y) = 8.69 m/s
look at your diagram and you will see
Tan = v(y)/v(x) = 8.69/1.31
= 81.4º
and
v(y) = v(i) sin
8.69 = v(i) sin(81.4)
v(i) = 8.79 m/s
so
clown B was launched at 8.79 m/s at 81.4º above horizontal

I have an equation that combines horizontal and vertical components and shortens calculation steps, however, because it shortcuts steps it leaves gaps in the logical process.

y = x*tan + 1/2*g* (x/(v(i)/cos )²
or
y = x*tan – 4.9* x² / v(i)² / cos²

= 81.43 degree

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