A 75.6-kg circus performer is fired from a cannon that is elevated at an angle of 37.2° 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 2.70 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 2.34 s to travel the horizontal distance of 26.9 m between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

The effective spring constant of the firing mechanism is approximately 21.804 N/m

Explanation:

The mass of the circus performer, m = 75.6 kg

The angle of elevation above the horizontal, θ = 37.2°

The instrument the cannon uses to propel the performer = Strong elastic band

The length by which the band stretches, x = 2.70 m

The height of the performer above the floor at the point the performer flies free of the bands = The height of the net into which he is shot

The time he takes to travel the horizontal distance between the point he is shot and the net, t = 2.34 s

The horizontal distance between the the point he is shot and the net, dₓ = 26.9 m

Therefore, the horizontal component of the velocity, vₓ, can be found with the following kinematic equation of motion;

vₓ = dₓ/t

∴ vₓ = 2.70 m/(2.34 s) = (15/13) m/s

By the resolution of the initial velocity, 'v', into the horizontal and vertical component, we have;

The horizontal component, vₓ = v × cos(θ)

∴ v = vₓ/(cos(θ))

v = (15/13)/(cos(37.2°)) ≈ 1.45

The initial velocity, v ≈ 1.45 m/s

The kinetic energy given to the circus performer by the elastic band, K.E., is given as follows;

K.E. = (1/2)·m·v²

Where;

m = The mass of the performer = 75.6 kg

v = The velocity given to the performer ≈ 1.45 m/s

∴ K.E. ≈ (1/2) × 75.6 × 1.45² ≈ 79.475

∴ K.E. ≈ 79.475 J

The kinetic energy, K.E., given to the performer, (1/2)×m×v² = The kinetic energy of the elastic band = (1/2)×k×x²

Where;

k = The effective spring constant of the firing mechanism

x = The extension of the sling = 2.70 m

Therefore, we get;

K.E. = (1/2)·m·v² = (1/2)·k·x²

∴ K.E. ≈ 79.475 J = (1/2) × 75.6 × 1.45² = (1/2) × k × (2.70 m)²

∴ 79.475 J ≈ (1/2) × k × (2.70 m)²

k = 79.475 J/((1/2) × (2.70 m)²) ≈ 21.804 N/m

The effective spring constant of the firing mechanism, k ≈ 21.804 N/m.

c, 1a hope this

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge