### Video Transcript

A sphere of mass 850 grams fell vertically from a height of 5.2 meters onto a horizontal section of the ground. It rebounded vertically upward. Given that the acceleration due to gravity is 9.8 meters per second squared and the loss in the sphereβs kinetic energy as a result of the collision is 2.54 joules, determine the maximum height the sphere reached after impact.

There is a lot of information in this question, so it is worth drawing a diagram first. Once we have done this, it is worth considering the steps we will need to take to obtain the final answer. We are told that the sphere falls from a height of 5.2 meters. It has a mass of 850 grams. As there are 1000 grams in a kilogram, we can rewrite this as 0.85 kilograms. The initial speed of the sphere is zero meters per second, and we will let the speed at which it hits the ground be π£ sub one. We also know that the acceleration due to gravity is 9.8 meters per second squared. We know that the height rebounds vertically upwards, and we will let the speed it rebounds be π£ sub two. We will let the maximum height that we are trying to calculate be β meters.

In order to calculate this maximum height, we will firstly consider the change in kinetic energy as we are told that the loss of the sphereβs kinetic energy as a result of the collision is 2.54 joules. We will then use the equations of motion or SUVAT equations to help us calculate π£ one and π£ two and then, following that, the height β. We know that the change in kinetic energy of a body is equal to the initial kinetic energy minus the final kinetic energy. The formula for calculating kinetic energy is equal to a half ππ£ squared, where the mass of the body is in kilograms, the velocity is in meters per second, and the kinetic energy will be given in joules.

As already mentioned, we know that the change in kinetic energy is 2.54 joules. The kinetic energy just before the collision with the ground is equal to a half multiplied by 0.85 multiplied by π£ sub one squared. And the kinetic energy after the collision is equal to a half multiplied by 0.85 multiplied by π£ sub two squared. We can divide both sides of the equation by a half and 0.85. This gives us 508 over 85 is equal to π£ sub one squared minus π£ sub two squared. As we have two unknowns here, we canβt go any further at this point, so we will call this equation one.

We will now use our equations of motion or SUVAT equations to consider what is happening when the sphere is descending. We know that the displacement π is equal to 5.2 meters, the initial speed π’ is equal to zero meters per second, the final velocity is equal to π£ sub one, and the acceleration is 9.8 meters per second squared. One of our equations states that π£ squared is equal to π’ squared plus two ππ . Substituting in our values, we have π£ sub one squared is equal to zero squared plus two multiplied by 9.8 multiplied by 5.2. The right-hand side here simplifies to 101.92. We can then square root both sides of this equation to calculate the velocity π£ sub one. This is equal to 10.0955 and so on.

Whilst this value would normally be important, if we look back at equation one, we can see that we have π£ sub one squared. This means that we can replace this with 101.92. This gives us 508 over 85 is equal to 101.92 minus π£ sub two squared. Adding π£ sub two squared and subtracting 508 over 85 to both sides gives us π£ sub two squared is equal to 95.9435 and so on. Once again, we can square root both sides of this equation to calculate π£ sub two. This is equal to 9.7950 and so on. It is important for accuracy that we do not round our answers until the end of the question. We will now clear some space and rewrite our values π£ sub two and π£ sub one as fractions.

We will now use our equations of motion once again to calculate the value of β. We know that at the maximum height, the velocity of the sphere will be equal to zero. Also, as it is now traveling upwards, the acceleration will be equal to negative 9.8 meters per second squared. We will once again use the equation π£ squared is equal to π’ squared plus two ππ . Substituting in our values gives us zero squared is equal to π£ sub two squared plus two multiplied by negative 9.8 multiplied by β. We know that π£ sub two squared is equal to 40776 over 425. Two multiplied by negative 9.8 multiplied by β is equal to negative 19.6β. We can add this to both sides of our equation. Finally, dividing both sides by 19.6 gives us a value of β equal to 4.89507 and so on. Rounding this to one decimal place gives us 4.9.

We can therefore conclude that the maximum height the sphere reaches after impact is 4.9 meters.