A circus performer is shot out of a cannon and flies over a net that is placed horizontally 5.3 m from the cannon. When the cannon is aimed at an angle of 30° above the horizontal, the performer is moving in the horizontal direction and just barely clears the net as he passes over it.
2007-06-13 15:59:46 · 3 answers · asked by terra_flare_aqua_ciel 1 in Science & Mathematics ➔ Physics
What is the muzzle speed of the cannon?
How high is the net?
2007-06-13 16:00:16 · update #1
((√3)/2)vt = 5.3
(1/2)v - 9.80662t = 0
v = 19.61324t
(√3)19.61324t^2 = 10.6
t^2 = 0.31203 s^2
t = 0.55860 s
v = 10.956 m/s
h = (10.956^2)/(4*19.61324)
h = 1.5300 m
2007-06-13 18:21:39 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
Muzzle velocity: 10.197 m/s
Height of net: 1.325 m
The flight of the performer is that of a parabola as can be seen by the equations of motion of the performer taken in the vertical and horizontal directions. Constant acceleration for vertical and constant velocity for horizontal.
The height of the net is the maximum height attained by the trajectory.
Knowing the equations of parabolic flight, the answer can be found.
2007-06-13 16:22:01 · answer #2 · answered by rahullak 1 · 0⤊ 0⤋
and the question is? ah ok... lets see
2007-06-13 16:05:34 · answer #3 · answered by driftaddict87 4 · 0⤊ 0⤋