A 68.6-kg circus performer is fired from a cannon that is elevated at an angle of 43.6° 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 3.09 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 the height of the net into which he is shot. He takes 1.92 s to travel the horizontal distance of 27.5 m between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

Question

A 68.6-kg circus performer is fired from a cannon that is elevated at an angle of 43.6° 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 3.09 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 the height of the net into which he is shot. He takes 1.92 s to travel the horizontal distance of 27.5 m between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

Answer in N/m

Answer 1

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Answer 2

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 given information.

1. The performer's mass (m) is 68.6 kg.
2. The angle of elevation (θ) of the cannon is 43.6°.
3. The elastic bands are stretched by a distance (x) of 3.09 m.
4. The horizontal distance (range) traveled by the performer is 27.5 m.
5. The time taken (t) to travel the horizontal distance is 1.92 s.

Now, let's proceed with the solution:

1. Find the initial velocity (v₀) of the performer in the horizontal direction:
v₀ = range / time = 27.5 m / 1.92 s

2. Find the vertical velocity (vᵥ) of the performer at the point where he flies free of the bands:
Since the performer is at the highest point in his trajectory at this point, his vertical velocity is 0 m/s.

3. Find the vertical displacement (h) of the performer from the point where he flies free of the bands to the height of the net:
The vertical motion can be analyzed using the equation y = v₀y * t + (1/2) * g * t², where v₀y is the vertical component of the initial velocity and g is the acceleration due to gravity.
Since the vertical velocity is 0 m/s, the equation simplifies to: h = (1/2) * g * t²

4. Find the vertical component of the initial velocity (v₀y):
v₀y = v₀ * sin(θ)

5. Find the gravitational acceleration (g):
g = 9.8 m/s²

6. Substitute the values into the equation for vertical displacement:
h = (1/2) * g * t² = (1/2) * 9.8 m/s² * (1.92 s)²

7. Since the performer's height above the floor is equal to the height of the net, the vertical displacement is equal to the stretched length of the elastic band (x):
x = h

8. Now, apply Hooke's Law to find the force (F) exerted by the stretched elastic band:
F = k * x, where k is the spring constant

9. Rearrange the equation to isolate k:
k = F / x

10. Substitute the force (F) with the weight of the performer (mg):
k = (m * g) / x

11. Substitute the known values and calculate the spring constant (k):
k = (68.6 kg * 9.8 m/s²) / 3.09 m

Now, let's perform the calculation:

k = (68.6 kg * 9.8 m/s²) / 3.09 m

k ≈ 219.3 N/m

Therefore, the effective spring constant of the firing mechanism is approximately 219.3 N/m.