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 = 76.0° angle, with a speed v0A = 9.50 m /s. The horizontal separation between the clowns as they leave the cannons is d = 5.55 m. Find the launch speed v0B and the launch angle B (>45.0°) for clown B.

Find Magnitude and direction.

Explanation / Answer

1) let 2 clowns meet after time t at position [x, 1.05] and [5.55 -x, 1.05]
angle of projection of 2nd is (p)

x = 9.5 cos 76 *t
5.55 - x = v0b cos (p) *t = v0b cos p *t
adding
v0b cos p = [5.55 - 9.5 cos 76 * t] / t ----------- (1)
--------------------
1.05 = 9.5 sin 76 *t - 0.5 gt^2
1.05 = vob sin p *t - 0.5 gt^2
--------------------------------
vob sin p = 9.5 sin 76 =9.21 ------------- (2)
===================== solving
1.05 = 9.5 sin 76 *t - 4.9t^2
4.9 t^2 -9.21 t + 1.05 = 0
t = 1.76 or t = 0.12
(1) gives
v0b cos p = 0.82 or v0b cos p = 43.95 ------------- (3)
two possibilities > whether collision took place when first was rising or falling at 1.05 m
----------------
from (1) & (3) squaring/adding
vob^2 = 9.21^2 + 0.82^2 >>> vob = 9.25 m/s
or
vob^2 = 9.21^2 + 43.95^2 >>> vob = 44.9 m/s
--------------------------------------...
directions> dividing equ 1 and 3
tan p = 9.21/0.82 >> p = 84.91 deg > 45 condition ----- (answer )
or
tan p = 9.21/43.95 >> p = 11.84 deg

which is less than 45 degrees so reject vob = 44.9 and tan p = 11.84 deg

so launch speed = 9.25m/s

launch angle = 84.91 deg

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