Please help me understand this physics question!

Question

Professional baseball pitchers deliver pitches that can reach the blazing speed of (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at  m/s . Assuming a pitched ball has a mass of  and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?

Recall that the acceleration due to gravity is . How high would the ball need to be dropped from to attain the same energy as it did when it was pitched (neglect air resistance)?

Answer 1

They want you to recall that kinetic energy is equal to 1/2 * mass * (the speed squared) = 1/2 m * v²

Which in this case would be 1 / 2 * .1436 kg * 41.34² (m²/s²) = 122.7 kg * (m²/s²) = 122.7 Joules

(It's important to understand that the definition of a joule of energy is the amount it takes to accelerate a 1 kilogram mass at a rate of 1 meter per second per second for a distance of one meter - it's a force times a distance. Knowing what each unit truly means makes physics MUCH easier)

Then they want you to take the amount of energy you just calculated, then recall potential energy from height converts to an equal amount of kinetic energy when you drop something, and potential energy due to height = mass * gravitational acceleration * height = mgh

Which in this case would mean 122.7 J (from our previous calculation) = .1436kg * 9.81 m/s² * h

So h = 87.1 meters

Getting to the point of really understanding this question means you have to understand energy. Energy, like momentum, is an idea based on Newton's Laws that we use to make problems simpler. We could figure out what height you have to drop a ball from to get a speed of 41.34 m/s with just Newton's Laws and the law of gravity (everything in the mechanics part of your physics class derives from them, so be sure you know them well), but energy lets us take a bit of a shortcut. So how is this accomplished?

We define work as a force times a distance it acts over (which is why Joules are equivalent to force units times distance units). If I accelerate some mass m at the rate due to gravity (g), then the force on it is mg. If I apply that force for a distance of h, then the work done is mg * h = mgh. So when we drop the ball, it gets mgh work done on it before it lands. We define energy based on work; the potential energy due to gravity that some object has is the amount of work gravity will do on it if we drop the object. The kinetic energy something has is the amount of work it takes to get the object to that speed (you can use Newton's Laws to calculate 1/2 * m * v² for the force applied * distance required to get the mass m to the speed v).

If that was a little too much, I'd say review your textbook's section on work, and see how the weight force mg times the fall distance h gives the work gravity will do on a released object. The insight that potential energy is just the work a gravitational force will do when applied over a fall distance can really help clarify why energy is so useful; we don't even need to know the exact path the falling object took!

Answer 2

K = (1/2)mv² = (1/2)(0.1436kg)(41.34m/s)² = 122.7059J

The 2nd part wants you to figure out the height required in order for the baseball to have the equivalent potential energy:

U = m*g*h = 122.7059J = (0.1436kg)*(9.81m/s²)*h

Solve for h: h = 87.105m

Answer 3

Kinetic energy is 1/2 times the mass times the velocity squared or

KE = 1/2mv²

So if m = 0.1436 kg ad v = 41.34 m/s then:

KE = 1/2(0.1436)(41.34²) = 122.7 joules

A ball placed at some height above the ground in preparation to be dropped would have gravitational potential energy equal to its mass times "g" times its height or:

PE = mgh

If we want to ensure the ball reaches the ground with the same kinetic energy as the pitched baseball, we would want to place it at a height so its gravitation potential energy would equal the kinetic energy of the pitched ball. So:

PE = 122.7 = mgh

122.7 = (0.1436)(9.81)(h)

h = 122.7/ (0.1436*9.81) = 87.1 meters