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.

Question

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.

Answer 1

To find the effective spring constant of the firing mechanism, we can use the principles of projectile motion.

Step 1: Find the velocity of the performer at the point where he flies free from the bands.
We can use the horizontal distance traveled and the time taken to find the horizontal component of velocity.
Given:
Horizontal distance, x = 24.7 m
Time taken, t = 2.72 s

Horizontal velocity, Vx = x / t = 24.7 m / 2.72 s = 9.088 m/s

Step 2: Find the vertical component of velocity at the point where the performer flies free.
Since the performer starts at ground level and reaches the same height as the net, the change in height is zero.
Using the kinematic equation:
Change in height = Vyi * t + 0.5 * g * t^2

The initial vertical velocity, Vyi, can be found using the equation:
Vyi = V * sin(angle), where V is the initial velocity and angle is the launch angle.

Given:
mass of performer, m = 65.0 kg
angle of elevation, θ = 55.8 degrees
stretch distance, d = 2.12 m

The initial potential energy stored in the elastic band is given by:
U = 0.5 * k * d^2

Since the performer initially rests on the bands, at the point where he is fired, all potential energy is converted to kinetic energy:
m * g * d = 0.5 * m * V^2

Simplifying,
g * d = 0.5 * V^2

Now, substituting the values:
9.8 m/s^2 * 2.12 m = 0.5 * V^2
V^2 = 41.9224
V = 6.4755 m/s

Now, we can find Vyi:
Vyi = V * sin(angle)
Vyi = 6.4755 m/s * sin(55.8 degrees)
Vyi = 6.4755 m/s * 0.8189
Vyi = 5.3017 m/s

Step 3: Find the initial vertical velocity (Vy) of the performer:
Since there is no vertical acceleration during the free-flight, Vy remains constant.
Vy = Vyi = 5.3017 m/s

Step 4: Calculate the spring constant (k) of the firing mechanism:
Using the horizontal and vertical components of initial velocity, we can calculate the total initial velocity (V).

V^2 = Vx^2 + Vy^2
V = √(Vx^2 + Vy^2)
V = √((9.088 m/s)^2 + (5.3017 m/s)^2)
V = √(82.692144 + 28.08938889)
V = √110.78153389
V = 10.523 m/s

Now, we can find the spring constant using the potential energy stored in the elastic band:
U = 0.5 * k * d^2

Rearranging the equation and substituting the known values:
k = 2 * U / d^2
k = 2 * (m * g * d) / d^2
k = 2 * (65.0 kg * 9.8 m/s^2 * 2.12 m) / (2.12 m)^2
k = 296.2264150943396 N/m

Therefore, the effective spring constant of the firing mechanism is approximately 296.23 N/m.

Answer 2

Horizontal range = v²sin2θ/g

=>
solve for v(initial).

Kinetic energy after launch = (1/2)mv²
Elastic potential energy before launch
=(1/2)kx²
Equate potential energies before and after launch to find k, the elastic constant of the mechanism.