If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
Given the data in the question;
Since the hockey puck was initially in the referee's hands
- Initial velocity;
![u = 0m/s](https://tex.z-dn.net/?f=u%20%3D%200m%2Fs)
- Distance or height from which it was dropped;
![h = 2.5m](https://tex.z-dn.net/?f=h%20%3D%202.5m)
- Acceleration due to gravity;
![g = 9.8 m/s^2](https://tex.z-dn.net/?f=g%20%3D%209.8%20m%2Fs%5E2)
- Coefficient of restitution a frozen puck;
![0.35](https://tex.z-dn.net/?f=0.35)
First we will find the velocity of the Puck when it hits the ground
From the Third Equation of Motion:
![v^2 = u^2 + 2as](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B%202as)
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
![v^2 = u^2 + 2gh](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B%202gh)
We substitute in our value and find "v"
![v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s](https://tex.z-dn.net/?f=v%5E2%20%3D%200%20%2B%20%282%20%5C%20%2A%5C%209.8m%2Fs%5E2%5C%20%2A%5C%202.5m%20%29%5C%5C%5C%5Cv%5E2%20%3D%2047.04m%5E2%2Fs%5E2%5C%5C%5C%5Cv%3D%20%5Csqrt%7B47.04m%5E2%2Fs%5E2%7D%5C%5C%5C%5Cv%20%3D%206.85857m%2Fs)
Now, Velocity of the hock puck after it hits the ground and bounce back;
We know that; Coefficient of restitution ![= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7BRelative%5C%20velocity%5C%20after%5C%20collision%7D%7BRelative%5C%20velocity%5C%20before%5C%20collision%7D)
Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision
we substitute in our values;
Relative Velocity after collision ![= 0.35 \ *\ 6.85857m/s](https://tex.z-dn.net/?f=%3D%200.35%20%5C%20%2A%5C%206.85857m%2Fs)
Relative Velocity after collision ![= 2.4 m/s](https://tex.z-dn.net/?f=%3D%202.4%20m%2Fs)
Now, to determine how high should the puck bounced back
We use the Third Equation of Motion:
![v^2 = u^2 + 2as](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B%202as)
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
![v^2 = u^2 + 2gh](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B%202gh)
Now, since the hockey puck bounces back, it is experiencing a negative acceleration
Hence, the equation becomes
![v^2 = u^2 - 2gh](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20-%202gh)
We substitute our values into the equation and find "h"
![(0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m](https://tex.z-dn.net/?f=%280m%2Fs%29%5E2%20%3D%20%282.4m%2Fs%29%5E2%20-%20%28%202%2A9.8m%2Fs%5E2%2Ah%29%5C%5C%5C%5C0%20%3D%205.76m%5E2%2Fs%5E2%20-%20%2819.6m%2Fs%5E2%2Ah%29%5C%5C%5C%5C%2819.6m%2Fs%2Ah%29%20%3D%20%205.76m%5E2%2Fs%5E2%20%5C%5C%5C%5Ch%3D%20%5Cfrac%7B%205.76m%5E2%2Fs%5E2%20%7D%7B19.6m%2Fs%5E2%7D%5C%5C%5C%5Ch%20%3D%200.3m)
Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
Learn more; brainly.com/question/16908019