Question

How do you do projectile motion problems

Answer 1

Here’s an example

Example 1

CSI discovers a car at the bottom of a 72 m cliff. How fast was the car going if it landed 22m horizontally from the cliff’s edge? (Note that the cliff is flat, i.e. the car came off the cliff horizontally).

v= ? [m/s]

Given: h=Δy=72 m

d=Δx=22 m

g=10.0 m/s2

Equation: h=viyt+12gt2 and d=vixt

Plug n’ Chug:

Step 1: Calculate the time required for the car to freefall from a height of 72 m.

h=viyt+12gt2 but since viy=0, the equation simplifies to h=12gt2 rearranging for the unknown variable, t, yields

t=2hg‾‾‾√=2(72 m)10.0 m/s2‾‾‾‾‾‾‾‾‾‾√=3.79 s

Step 2: Solve for initial velocity:

vix=dt=22 m3.79 s=5.80 m/s

The answer is 5.80 m/s.

Answer 2

The key  projectile motion is that gravity allows downward only