BlameThePixel!

^sup and _sub tags.

Alright, I know super/subscript now, time for a simple problem.

Joey shot a cannon. The cannon was pointing 60^o from the horizontal, and fired with a velocity of 10m/s. Directly in the middle of the ball's path, sat a wall 5m high. Will the ball clear the wall, and where will the ball hit (anything)? Wind resistance is negligent and should be ignored.

Step 1: Find total velocity in x and y directions.

sin theta = V_y/r
sin 60^o = V_y/10
0.8660 = V_y/10
8.66m/s = V_y

cos theta = V_x/r
cos 60^o = V_x/10
0.5 = V_x/10
5m/s = V_x

Step 2: Find the total time.
Reasoning: Y max is positive at the firing point, and Y max is negative at the landing point. Acceleration is equal to negative gravity

V_yf = V_yi + at
-8.66 = 8.66 + (-9.81)t
-17.32 = (-9.81)t
1.77s = t

Step 3: Find max height in y axis.
Reasoning: Maximum height must be in the exact center if wind resistance is disregarded. Thus, V_yf should be set to zero.

V_yf² = V_yi² + 2ay
0 = 8.66² + 2(-9.81)y
0 = 74.9956 + (-19.62)y
-74.9956 = -19.62y
3.82m = y

The ball did not clear the wall.

Step 4: Find the distance to the ball.
Reasoning: Since the ball stopped when it hit the wall, the distance of the ball and the distance to the wall are equal. Since the wall must be right in the middle, time should be cut in half. Since there is no wind resistance, the balls initial and final velocity are the same.

x = ([V_xi V_xf] / 2)t
x = ([5.0 + 5.0] / 2)(0.855)
x = (10 / 2)(0.855)
x = 5(0.855)
x = 4.28m

The ball flew 4.28 meters before it hit the wall.

And there you have it, a simple physics problem.

That's about the maximum capacity of what we were learning in physics...

Same here, I dunno what A levels gonna be like

The last experiment you gave us, the burning matches is to do with chemistry. But most of it is Physics, im just being piccy.

I'll just give you the formula we started using before the weekend, since I still don't fully understand how to do the problem.

P₁ + (0.5)dv₁² + dgh₁ = P₂ + (0.5)dv₂² + dgh₂

I'm afraid you lost me at sin theta...

theta is the greek letter used for the angle. sin theta is finding the sine of the angle.