I'm Desperate!

[ruferddude]

I have to write an essay for physics about how playing basketball on the moon would be different. In the rubric he says to include the following information in the report:

4. Calculate how far a ball will travel horizontally on the Earth, if a player throws the ball at an angle of 45 degrees above the horizontal with an initial velocity of 12.23 m/s, and another player catches it at the same height as it was thrown. Do the same calculation for the Moon.

I know the gravitational acceleration for Earth is -9.81m/s^2 and the Moon is -1.67m/s^2. It never gives how long the ball is in the air. I can't figure out the displacement without knowing the change in time (or "delta T") and I can't find the change in time without the displacement. I need help please!

[Flyleef]

Re: I'm Desperate!

Let:
u be the initial velocity,
v be the final velocity,
s be the distance fallen,
g be the acceleration due to gravity.

The equation of motion is:
v^2 = u^2 + 2gs.

As the baall is initially at rest, u = 0.
Therefore:
v^2 = 2gs.

On earth:
v^2 = 2 * 9.81 * 1.5
v = sqrt(2 * 9.81 * 1.5)
= 5.42 m/s.

On the moon:
v^2 = 2 * 1.6 * 1.5
v = sqrt(2 * 1.6 * 1.5)
= 2.19 m/s.

4.
Let:
x be the horizontal distance travelled,
t be the time of flight,
a be the angle to the horizontal at which the ball is projected,
u be the initial velocity,
g be the acceleration due to gravity.

Resolving vertically and horizontally:
0 = tu sin(a) - gt^2 / 2 ...(1)
x = tu cos(a) ...(2)

From (1):
u sin(a) = gt / 2
t = 2u sin(a) / g

Substituting this value of t in (2):
x = 2u^2 sin(a)cos(a) / g
= u^2 sin(2a) / g.

On earth:
x = 12.23^2 * 1 / 9.81
= 15.24m.

On the moon:
x = 12.23^2 * 1 / 1.6
= 93.48m.

- Yahoo

[ruferddude]

Re: I'm Desperate!

I tried looking on yahoo but couldn't find anything. I was just probably looking for it the wrong way. Man I feel dumb. Anyways, thank you for helping me.

[Flyleef]

Re: I'm Desperate!

No problem. And remember that everything about life is on the internet; search engine. ^______^;