A 65.0-kg circus performer is fired from a cannon that is elevated at an angle of 55.8 ¡ã above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by 2.12 m from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as that of the net into which he is shot. He takes 2.72 s to travel the horizontal distance of 24.7 m between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

LOL, the way MathMate did it is a lot easier :)

See earlier post:

http://www.jiskha.com/display.cgi?id=1400921781

To determine the effective spring constant of the firing mechanism, we can use the principles of projectile motion and energy conservation.

First, let's analyze the motion of the performer in the horizontal and vertical directions separately.

In the horizontal direction:
The time taken by the performer to travel a horizontal distance of 24.7 m is given as 2.72 s.

Using the equation for horizontal velocity: v = d/t, where v is the horizontal velocity, d is the horizontal distance, and t is the time, we can calculate the horizontal velocity (v) of the performer:
v = 24.7 m / 2.72 s
v ≈ 9.095 m/s

In the vertical direction:
The performer is launched at an angle of 55.8 degrees above the horizontal. The initial vertical velocity (v₀y) can be found using the equation: v₀y = v₀ * sin(θ), where v₀ is the initial velocity and θ is the launch angle.

Using the given information, we know that the initial vertical displacement (y) of the performer is equal to the stretched length of the bands, which is 2.12 m.

Using the equation for vertical displacement: y = v₀y * t + (1/2) * g * t², where g is the acceleration due to gravity and t is the time, we can solve for v₀y:
2.12 m = v₀y * 2.72 s + (0.5) * 9.8 m/s² * (2.72 s)²
v₀y ≈ 7.376 m/s

Now, let's use the concept of conservation of mechanical energy to find the spring constant (k) of the firing mechanism.

The initial mechanical energy is stored as potential energy in the stretched elastic bands:
E₀ = (1/2) * k * x², where k is the spring constant and x is the stretched length of the bands (2.12 m in this case).

The final mechanical energy is converted entirely into the kinetic energy of the performer as he leaves the bands:
E = (1/2) * m * v², where m is the mass of the performer (65.0 kg) and v is his velocity (9.095 m/s horizontally).

Equating the initial and final mechanical energies, we have:
(1/2) * k * x² = (1/2) * m * v²
k * x² = m * v²
k = (m * v²) / x²

Plugging in the values we know, we can solve for k:
k = (65.0 kg * (9.095 m/s)²) / (2.12 m)²

Evaluating this expression, the effective spring constant of the firing mechanism is approximately 444 N/m.

The total energy at the top of the trajectory is the same as was stored in the spring (1/2)kx^2

Total energy at top where v = 0 = (1/2) u^2 + m g h

we can find u, horizontal velocity easily
u = 24.7 m/2.72 s = 9.08 m/s

but Vi, initial speed up is easy too
tan 55.8 = Vi/u
Vi = 9.08 * tan 55.8 = 13.4 m/s

so we need to find where v = 0 to get the height
v = Vi - 9.81 t
0 = 13.4 - 9.81 t
t = 1.36 seconds to top
(which is 2.72 / 2 of course. It goes up half the time :)

so how high in 1.36 seconds?
h = Vi t - 4.p t^2
= 13.4(1.36) - 4.9 (1.36^2)
= 18.22 - 9.06 = 9.1 meters high at top
potential energy at top = m g h
= 65 * 9.81 * 9.1 = 5802 Joules
kinetic energy at top = (1/2)65(9.08^2)
= 2680
total energy = 8480 Joules
(1/2) k 2.12^2 = 8480
k = 3774 N/m