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., the x 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 the x 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