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## 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.