The AMOUNT of energy the ball has doesn't change. It's 294 joules in Darwin's hand, and it's still 294 joules when the ball hits the ground. It's all PE before he let's it go, and it steadily changes from PE to KE all the way down.
It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.
More and more PE turns into KE as the ball falls, all the way down.
When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
The hotter an object is, the more kinetic energy it has, but i'm not sure what is the exact word missing??
Potential energy (PE ) = m g h
Where:
m = mass = 3800 kg
g = acceleration due gravity = 10 m/s^2
h = heigth = 110 meters
Replacing:
PE = 3800 * 10 * 110 = 4,180,000 J
