Question

determining the potential and kinetic energy of a ball
read the scenario, then answer the questions.
a 2 kg ball is thrown upward with a velocity of 15 m/s.
what is the kinetic energy of the ball as it is being thrown? ◇ j
what is the potential energy of the ball when it gets to its maximum height just before falling back to the ground?
◇ j

Explanation:

First Question: Kinetic Energy of the Ball as it is Being Thrown

Step1: Recall the formula for kinetic energy

The formula for kinetic energy (KE) is $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the object and $v$ is its velocity.

Step2: Identify the given values

We are given that the mass $m = 2\space kg$ and the velocity $v = 15\space m/s$.

Step3: Substitute the values into the formula

Substitute $m = 2$ and $v = 15$ into the formula:
$KE=\frac{1}{2}\times2\times(15)^2$
First, calculate $(15)^2 = 225$. Then, $\frac{1}{2}\times2 = 1$. So, $KE = 1\times225 = 225\space J$.

Second Question: Potential Energy at Maximum Height

Step1: Recall the principle of conservation of mechanical energy

At the maximum height, the kinetic energy of the ball is zero (since its velocity is zero at the peak), and all the initial kinetic energy is converted into potential energy (assuming no air resistance). So, the potential energy (PE) at maximum height is equal to the initial kinetic energy.

Step2: Use the result from the first question

From the first question, we found that the initial kinetic energy is $225\space J$. Therefore, the potential energy at maximum height is also $225\space J$ (by conservation of mechanical energy, $KE_{initial}=PE_{maximum}$).

Answer:

s:

• Kinetic energy as it is being thrown: $\boldsymbol{225}$ J
• Potential energy at maximum height: $\boldsymbol{225}$ J