D. The osculations show a variable rate of motion. Hope this helps:)
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 answer is A ..........
I would tell him, in the kindest, most gentle way I could manage,
to fahgeddaboudit.
The total amount of energy doesn't change. Energy is never created,
and it never disappears. If you have some energy, then it had to come
from somewhere, and if you used some energy, then it had to go
somewhere.
You can never get more energy out of the electromotor than you put into it,
and in the real world, you can't even get THAT much out, because some
of it is always used on the way through.
Pour yourself a cold glass of soda, then look up "Perpetual Motion" or
"Free Energy" on the internet, relax, and enjoy the show. They are all
fakes. They may not all be intentionally meant to fool you, but they are
all impossible.
