Question

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

Answer 1

Answer:

Lets find the maximum height first at max height (h) final velocity (Vf) would be 0, ‘g’ would be negative because its going upwards against the gravity and since its Vf would be 0 means it is decelerating

2*a*S= Vf^2 - Vi^2

or 2*-g*h= Vf^2 - Vi^2

2*-9.8*h= 0 - 15^2

-19.6*h= -225

h= 225/ 19.6

h= 11.48 m

Kinetic Energy

$= \frac{1}{2} m {v}^{2} = \frac{1}{2} \times 2 \times {15}^{2} = 225 \: j$

Potential Energy

$= mgh = 2 \times-9.8 \times 11.48 \\ = - 225.008 \: j$

Answer 2

Answer:

They both are 225.

Explanation: