Week of May 6th Physics Problem: Determining the Range and Maximum Height of a Projectile

In a student's first semester of college physics and after learning the fundamental equations governing the displacement, velocity, and acceleration of an object with respect to one dimension (e.g., thex direction), students oftentimes have a little more difficulty once two dimensional motion is considered. A typical problem which incorporates both x and y movement is one which involves the trajectory of a projectile. For example:

Suppose a cannonball is fired at an angle of 30° (relative to the horizontal) at an initial velocity of 100 m/s. What is the range and maximum height of the projectile? (Ignore air resistance)

To calculate the range (or horizontal distance traveled) of the cannonball, we need to calculate thex and y components of the initial velocity and the time the projectile is in the air. Thus,

velocity(x) = velocity(initial) (cosine θ) = 100 m/s (cosine 30) = 86.6 m/s

velocity(y) = velocity(initial) (sine θ) = 100 m/s (sine 30) = 50.0 m/s

time in air = [2 (velocity initial) (sine θ)] / g = [2 (100 m/s) (0.5)] / 9.81 m/s^2 = 10.2 s  

Using these data, we calculate the range as follows:

Range = (velocity x) (time) = (86.6 m/s) (10.2 s) = 883 m

Maximum height = (velocity y) (½time in air) - [½ g (½time in air)^2] = (50.0 m/s) (5.1 s) - [(½ (9.81 m/s^2) (5.1 m/s)^2] = 128 m

Note that only half the time in the air was used (5.1 s) because the projectile traces out a parabolic path, with its maximum height achieved after it has traveled halfway to its destination (due to the mirror symmetry of the parabola). For a great video: https://www.youtube.com/watch?v=Q53HHMMWtf0

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