What is the relationship between range and initial height in projectile motion?

AI Thread Summary
The relationship between range and initial height in projectile motion is complex and depends on various factors, including initial velocity and launch angle. Doubling the initial height does not yield a straightforward answer without knowing how other parameters change, such as velocity or angle. The range is calculated using the horizontal velocity and time of flight, which are influenced by the height of the projectile. The discussion highlights that while increasing height can affect range, the specific impact requires a detailed analysis of the projectile's initial conditions. Understanding these dynamics is crucial for accurately predicting the maximum horizontal distance a projectile can travel.
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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?
 
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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?
 


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
 


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?
 


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.
 
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