A 70-kg circus performer is fired from a cannon that is elevated at an angle of 35° above the horizontal.

Question

A 70-kg circus performer is fired from a cannon that is elevated at an angle of 35° 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 3.3 m from their unstrained length. He takes 4 s to travel between the launch point (where he is free from the bands) and the net into which he is shot. Assume the launch and landing points are at the same height and do not neglect the change in height during stretching.

a) What is the launch speed? (Ignore friction and air resistance.) ______m/s

b) What is the effective spring constant of the firing mechanism? ______N/m

Answer 1

Question: A 70-kg circus performer is fired from a cannon that is elevated at an angle of 35° 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 3.3 m from their unstrained length. He takes 4 s to travel between the launch point (where he is free from the bands) and the net into which he is shot. Assume the launch and landing points are at the same height and do not neglect the change in height during stretching.

a) What is the launch speed? (Ignore friction and air resistance.) _34 m/s_

b) What is the effective spring constant of the firing mechanism? _7,430_N/m

Solution

His total flight time is 4 seconds. We know from Projectile motion that max height is reached in 1/2 the total flight time; in this case, that would be 2 seconds.

Using the expression for distance traveled under uniform acceleration: y = 1/2 at² with no initial velocity and using 9.8 m/sec² (gravity) for the acceleration and the 2 seconds to max height, the performer reached a height of y = 1/2 (9.8 m/sec²)*4 = 19.6 meters. This will allow us to compute his initial velocity using the expression for uniform acceleration knowing only the distance and acceleration: V_f² - V_i² = 2ay. We also know that at max height, the performer's final vertical velocity is zero; thus V_i² = 2ay = V_i² = 2*(9.8)*19.6, V_i = 19.6 m/s. Their initial vertical velocity, V_iy = 19.6 m/s. Identifying the resultant using trig functions: V_i = V_iy/sin 35 = 34 m/s.

Equating the potential energy of the elastic bands (modeled as springs) to the kinetic energy of the performer at launch time will allow us to compute the "spring constant" k: 1/2kx² = 1/2mv². Using the value for their initial launch velocity of 34 m/s, the performer's mass of 70 kg, the stretch length of the bands at launch (3.3 m) and solving for k, the spring constant, we get 7,430 N/m as the spring constant.