Projectile Motion Example - Foundations of Physics | PHYS 1000, Assignments of Physics

Download Projectile Motion Example - Foundations of Physics | PHYS 1000 and more Assignments Physics in PDF only on Docsity! Projectile motion example: An object is launched into the air with an initial horizontal speed of 8 m/s and an initial vertical speed of 30 m/s. Find the horizontal distance traveled by the object. To solve, decompose the motion into its horizontal and vertical components. Then, proceed to solve each one, using time at the link between the two components of the motion. Given information: Vhor (initial) = 8 m/s Vver (initial) = 30 m/s Known information: ahor = 0 m/s2 aver = -g = -10 m/s2 Also, for all vertical motion problems, it is known that the vertical velocity at the maximum height is 0 m/s. Vertical motion: In general, the vertical motion can often (but not always) be used to determine the time. The initial vertical velocity is directly tied to the time a projectile spends in the air. So, first, find the time to get to the maximum height: Use v = at t = v/a = (-30 m/s) / (-10 m/s2) = 3 sec [Note, -30 m/s is used because the equation assumes STARTING from rest and reaching some final velocity. We have reversed the situation here.] But this is just the time to get to the maximum height. The total time in the air is TWICE this value (time up + time down), assuming that you start and end at the same height. Therefore, the total time in the air is 6 seconds. Horizontal Motion: Horizontal travel has no acceleration, so: d = vt (using total time) d = v*t = (8 m/s) * (6 s) = 48 m