# Range and Height Kinematics?

1. Oct 22, 2012

### pxp004

Range and Height Kinematics????

What happens to range if the initial height is doubled. Say if something is shot at 2 ft and then its shot at 4 ft high, what happens to range?

2. Oct 22, 2012

### Simon Bridge

Re: Range and Height Kinematics????

That is a good question - not enough information to have a unique answer though. eg - how was the height doubled? Are we thinking that the initial velocity is increased but the angle is the same, that the initial velocity is the same but the angle changes, that the y-component of the initial velocity is increased, or that the launcher is raised, everything else remains the same?

3. Oct 22, 2012

### pxp004

Re: Range and Height Kinematics????

a projectile is shot at an angle with a certain initial velocity at a height, say 2 ft above ground. The only thing that changes is the height above the ground

4. Oct 22, 2012

### Simon Bridge

Re: Range and Height Kinematics????

OK - so you need to know how initial height affects the range!

Note: $R=v_xT$ - the range is equal to the time-of-flight T times the horizontal velocity. Which of these variables changes with height?

5. Oct 23, 2012

### haruspex

Re: Range and Height Kinematics????

If by range you mean the maximum horizontal distance, you have:
- initial speed u
- initial trajectory θ to horizontal
- height above target h
- flight time t
- horizontal range r
Rather than compute the max r for given h, turns out to be a little easier to ask for min h for given r:
r = u cos(θ) t
h = - u sin(θ) t + g t2 / 2
= - r tan(θ) + g r2 sec2(θ)/ 2 u2
dh/dθ = -r sec2(θ) + g r2 tan(θ) sec2(θ)/u2
For min h, dh/dθ = 0
tan(θ) = u2 /(gr)
h = g r2 / 2 u2 - u2 /2g
Looks a little strange, but I think it's right. E.g. if the target is u2 /2g above the launch then h = - u2 /2g and r = 0.