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

See, they collide at 1m above the surface. This means, both of 'em traveled 1m above the surface. Since both of them have same vertical speed, it makes things easier.

Vertical Distance = ut-1/2at^2 = 7.65t-4.9t^2 = 1 m.

Both will be at a height of 1 m twice, once during ascent, and once during descent.

So the variable 't' will have two solutions, 1.417 seconds and 0.144 seconds.

Now, since they meet/collide, the sum of horizontal distance traveled by both has to be 6 m.

v0Ax*t + v0Bx*t = 6

2.34t + vt = 6 ----Eq 1

When they collide, two conditions are simultaneously satisfied. 1) They are both at 1 m height. 2) The sum of their distances from initial positions is 6 m. Since both events occur at the same time, the time can be either 1.417 s or 0.144 s since at both these times, they are 1 m above the ground.

Now,

tan B = VyB/VxB

But B > 45. => tan B > 1 => V0yB/V0xB > 1 => V0yB > V0xB =>V0xB < 7.65 m/s === Eq 2

Using t=0.144 s in equation 1, we get V0xB=39.32 m/s which can't be true as proved above.
Using t=1.417 s in the equation, we get V0xB= 1.9 m/s which is consistent with Eq 2.

Now tan B = 7.65/1.9 ~ 4 => B ~ 76 degrees.