Relationship between Launch height and range of a projectile

  • #1

Homework Statement


In a practical that we conducted, we were investigating the relationship between Launch height and the range (horizontal distance it travels) of a projectile. We launched a projectile from various heights and measured the horizontal range it travelled before reaching the ground. We concluded that as height increases, range also increases. We are now trying to explain why an increase in launch means an increase in range.

2. The attempt at a solution
This is our attempt at explaining why this occurs.

"The results suggest that an increase in launch height causes an increase in range, as was predicted in the hypothesise. This relation can be linked directly to the physics concepts of projectile motion. Range depends upon two factors, horizontal velocity and air time of the projectile. Since velocity was constant throughout the investigation, the only factor that had an impact upon the range is air time. Increasing the height means that the projectile has a greater vertical distance to travel. The upwards motion of the projectile is unchanged regardless of the height, however the difference in range between each height occurs due to the increased or decreased downwards distance the projectile must travel which is the result of the launch height. Increasing the launch height increases the downward distance, giving the horizontal component of the velocity greater time to act upon the projectile and hence increasing the range. "

Also, we derived an equation for range (S) in terms of height (h).

S = (vcostheta(vsintheta + (v^2sin^2theta-2gh)^1/2)/-g

where v = initial velocity, theta = launch angle, and g=gravity

This is the equation in terms of the values that we got in our practical. v = 2.94 , theta = 45 degrees, g = -9.8

therefore , S = (2.94cos45)(2.94^2(sin45)^2-2(-9.8)(h))^1/2)/-(-9.8)

This maybe useful.

We were looking for a more scientific explanation which could maybe include mathematical explanations as well, however we could not create one ourselves. We couldn't find anything worthwhile on the internet either. It will be really useful if someone can help us out.
 

Answers and Replies

  • #2
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I think the explanation is fine. You can simplify it, however: Neglecting things like the air density changing with height (irrelevant at your parameters), increasing the launch height is equivalent to lowering the floor. The mass will always have the same horizontal distance when it crosses the height of the launch point again. Afterwards it continues to travel forwards, and the more height it has at this point (the lower the floor is relative to this point) the more it will travel forwards before hitting the ground.
 
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  • #3
kuruman
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I agree with the assessment by @mfb. However, I feel obliged to make a remark involving good habits that will serve you well in the future, especially when you draw and label force diagrams. Think of g as the magnitude of the acceleration of gravity, that is g = +9.8 m/s2. For projectile motion you start with the general equation for constant vertical acceleration y(t) = y0 + v0y t +(½)ay t2 and then adapt it tot he specific problem. Here, y0 = h, v0y t = v0sin(θ) and ay = -g. Therefore, y(t) = y0 + v0sin(θ) t - (½)g t2 where g is a positive quantity. Note how the component of the acceleration ay is given a negative algebraic sign indicating that the direction is "down". No more negative signs need to be introduced as you proceed with the algebra and, when the time comes to substitute numbers, you replace g with +9.8 m/s2.
 

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