Question Video: Finding the Maximum Height a Vertically Falling Ball Reaches after Rebounding from the Ground

A ball of mass 100 g fell vertically from a height of 7 m onto a section of horizontal ground. The ball hit the ground and rebounded vertically upward. The loss of kinetic energy as a result of the collision was 1.568 joules. Determine the maximum height the ball reached after the first bounce. Consider the acceleration due to gravity to be 𝑔 = 9.8 m/sΒ².

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Video Transcript

A ball of mass 100 grams fell vertically from a height of seven meters onto a section of horizontal ground. The ball hit the ground and rebounded vertically upward. The loss of kinetic energy as a result of the collision was 1.568 joules. Determine the maximum height the ball reached after the first bounce. Consider the acceleration due to gravity to be 𝑔 equal to 9.8 meters per second squared.

In this question, we need to consider the kinetic energy of the ball. We know that kinetic energy is equal to a half π‘šπ‘£ squared, where π‘š is the mass and 𝑣 the velocity or speed. When the mass is measured in kilograms and velocity in meters per second, the resulting units for the energy will be joules. We are told that the mass of the ball is 100 grams. And as there are 1000 grams in a kilogram, this is equal to 0.1 kilograms.

Sketching the situation, we see that the ball is dropped from a height of seven meters. Its initial speed 𝑒 will be zero meters per second. The acceleration due to gravity is 9.8 meters per second squared. We will let the velocity that the ball hits the ground be 𝑣 one and the velocity that it rebounds be 𝑣 two. We will call the maximum height which we’re trying to determine β„Ž. In order to calculate some of these unknowns, we’ll use our equations of motion or SUVAT equations, where 𝑠 is the displacement or distance, 𝑒 is the initial velocity, 𝑣 is the final velocity, π‘Ž is the acceleration, and 𝑑 is the time.

Let’s firstly consider the motion of the ball when it is dropped. If we consider the positive direction to be downward, we know that 𝑠 is equal to seven meters, 𝑒 is zero meters per second, 𝑣 is equal to 𝑣 one, the velocity on impact, and π‘Ž is equal to 9.8 meters per second squared. We can use the equation 𝑣 squared is equal to 𝑒 squared plus two π‘Žπ‘  to calculate 𝑣 one. Substituting in our values, we have 𝑣 sub one squared is equal to zero squared plus two multiplied by 9.8 multiplied by seven.

The right-hand side simplifies to 137.2. We can then square root both sides of this equation so that 𝑣 sub one is equal to seven root 70 over five. The velocity at which the body strikes the ground is seven root 70 over five meters per second. We can now calculate the kinetic energy at this point. The kinetic energy is equal to a half multiplied by 0.1 multiplied by seven root 70 over five squared. This is equal to 6.86 joules.

We are told that, as a result of the collision, the loss in kinetic energy is equal to 1.568 joules. This means that we can calculate the kinetic energy after impact by subtracting this from 6.86. This gives us 5.292 joules. We can now use this value together with the kinetic energy formula to calculate the value of 𝑣 sub two, the velocity of the ball after the collision. 5.292 is equal to a half multiplied by 0.1 multiplied by 𝑣 sub two squared. We can multiply both sides of this equation by two and then divide by 0.1 such that 𝑣 sub two squared is equal to 105.84. Square rooting both sides of this equation gives us 𝑣 sub two is equal to 21 root six over five. The velocity of the ball after the collision is 21 root six over five meters per second.

We can now use the equations of motion once again to calculate the value of β„Ž. Doing this motion, the ball is traveling upward. And we will consider this to be the positive direction. The initial velocity here is 21 root six over five meters per second. We know that at the maximum height the velocity will be zero meters per second. Therefore, 𝑣 equals zero. As gravity is acting against the motion, π‘Ž is equal to negative 9.8 meters per second squared.

We are trying to calculate the height β„Ž. Substituting in these values into 𝑣 squared is equal to 𝑒 squared plus two π‘Žπ‘  gives us zero squared is equal to 21 root six over five squared plus two multiplied by negative 9.8 multiplied by β„Ž. This simplifies to zero is equal to 105.84 minus 19.6β„Ž. We can add 19.6β„Ž to both sides. Dividing both sides of this equation by 19.6 gives us β„Ž is equal to 5.4.

The maximum height that the ball reached after the first bounce is 5.4 meters.

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