does anyone know where I can find help on a physics question i have?

Question

the question is "How does the horizontal distance travelled by a ball projected from a ramp vary with the height of the ramp?"

Answer 1

I assume you are shooting your projectile from a ramp that is inclined (like an archer loosing off his arrow , he points the arrow up at an angle before letting go).
Bear with me on this, for it's hard to do without a diagram. You'll have to think hard about angles, and draw them out for yourself!
You'll also have to be comfortable with uniform motion in a straight line, and in particular, the formula v = u + at. Finally, you'll need to be able to resolve velocity into two mutually perpendicular components
Suppose the ramp is inclined at an angle A to the horizontal.
(Draw this, marking in the angle A- and also note the angle between the ramp and the VERTICAL. This step is important to your understanding).
Suppose now that the projectile is released with a velocity V.
There are two components of this velocity, one in the horizontal, and one in the vertical.
Looking at your diagram, you should be able to work out that the initial velocity in the horizontal is VcosA.
In the vertical, the initial velocity will be VsinA.
The velocity in the vertical will be affected by gravity,
What's done now is to work out how long in time the projectile will travel vertically. It's going straight up, then stops and falls back down.
The formula v=u+at is taken and is used for the upward part as follows:
v is the final velocity
u is the initial velocity
a is the acceleration, in this case gravity,-g (it's - because the acceleration is downwards, but the motion is upwards)
t is the time taken to reach final velocity from the start.
In the case of our projectile,
u, the initial velocity, is VSinA
v, the final velocity is 0 (the projectile stops going up)
Plugging these values into the equation, we see
0 = VSinA -gt
so the time t taken to reach the maximum height is
VSinA/g
The time taken for the whole flight (up and down!) is twice this,
namely 2VSinA/g
Here's the cunning bit- this time will be the same time as that spent going horizontally.
For horizontal motion, gravity has no effect, so the initial velocity (VCosA, remember?) will stay the same throughout.
When the projectile lands, it will have travelled for a time t at a constant speed of VCosA.
So the distance gone horizontally will be the speed times the time.
If this distance is R, then R=VCosA times 2VSinA/g
Thus R=2V^2CosASinA/g.
This simplifies to
R = V^2 Sin2A/g, since 2SinACosA = Sin2A

All this assumes that air resistance is zero, that the ground itself is horizontal down range, and that g does not vary.
You should be able to see that
1. If the speed of the projectile is doubled, the range quadruples, being proportional to V^2
2. The range is a maximum if Sin2A is a maximum, and this occurs when 2A = 90 degrees, or A=45 degrees
3. The range is inversely proportional to g, so the projectile would go 6 times as far on the moon, whose gravity is 1/6 of that on the Earth.
Good luck with this!

Answer 2

The higher the ramp, the further the distance.

Answer 3

A higher ramp will impart more velocity to the ball.

Answer 4

Try www.gcse.com/physics