The above is not complete. Couldn't find the it online but i have found a similar question which would help you solve the above
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 = [tex]\frac{Distance}{time} =\frac{26.8}{2.14} = 12.523 \ m/s[/tex]
To obtain the initial we would divide the final velocity by the cosine of the angle of elevation
Initial velocity = [tex]\frac{12.523}{cos\ 40} = 16.348m/s[/tex]
Next is to obtain the initial kinetic energy
This is equal to = [tex]\frac{1}{2} mv^2= \frac{1}{2} *70*16.348^2 = 9354 \ Joules[/tex]
Looking at the diagram in the second uploaded the cannon raised the performer before releasing him is [tex]30sin(40) = 19284 \ m[/tex]
So the potential energy given by the cannon is = mgh [tex]=70 *9.80*1.9284 = 1323 \ Joules[/tex]
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 [tex]=\frac{1}{2}kx^2[/tex]
And we have calculated the stored energy as 10677
Substituting
[tex]10677 = \frac{1}{2} *k * 3.00^2[/tex]
[tex]k = 2373 \ N/m[/tex]
