Question

A 70-kg circus performer is fired from a cannon that is elevated at an angle of 37° 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 3.8 m from their unstrained length. He takes 4.1 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.

Answer 1

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Answer:

Explanation:

From the question we are given that

Mass = 70kg

Angle of elevation = 40°

Length of  stretch = 3.00 m

Height of performer above floor = Height of the net into the into which he is shot

  Time traveled = 2.14 s

Distance traveled = 26.8 m

First obtain the horizontal velocity = $\frac{Distance}{time} =\frac{26.8}{2.14} = 12.523 \ m/s$

To obtain the initial we would divide the final velocity by the cosine of the angle of elevation

     Initial velocity = $\frac{12.523}{cos\ 40} = 16.348m/s$

Next is to obtain the initial kinetic energy

This is equal to = $\frac{1}{2} mv^2= \frac{1}{2} *70*16.348^2 = 9354 \ Joules$

Looking at the diagram in the second uploaded the cannon raised the performer before releasing him is $30sin(40) = 19284 \ m$

So the potential energy given by the cannon is = mgh $=70 *9.80*1.9284 = 1323 \ Joules$

 Hence the total energy the band gives the performer i.e the total energy stored in the band is  = 9354 +1323 = 10677 Joules

        To obtain the Spring constant we would use the stored energy formula

       i.e  Stored Energy $=\frac{1}{2}kx^2$

And we have calculated the stored energy as 10677

      Substituting

               $10677 = \frac{1}{2} *k * 3.00^2$

                     $k = 2373 \ N/m$