Forces and Motion

Equation

y = viyt + ½ gt2

1.13

Equation

x = vixt + ½ axt2

1.11

Projectile Motion

Understanding Projectile Motion

Here is an example of using these equations to understand projectile motion.

Scenario:

A cannon on top of a cliff facing a river is fired at an attacking warship 100 m below. The ship is 400 m off shore, so at what velocity does the cannonball have to be fired to hit the enemy?

The way to solve most problems is to make a drawing of the distances and forces, and to list what is known and what needs to be determined.

Known:

• Height difference (y) between cannon and ship = 100 m

• Horizontal distance to ship = 400 m

• Cannon ball fired horizontally and gravity pulls ball downward

Want to calculate: Velocity necessary for cannonball to hit ship.

First, we find out how long (t) a cannonball takes to fall 100 m under the pull of gravity.

Use Eqn 13 above:

y = viyt + ½ gt2

This is our problem and we can do anything (legal) to make it easier to solve, so we move the origin of the x-y coordinate axis to the cannon, rather than on the bottom left of the diagram, and we make down the positive y direction, because that is the way the cannonball will move. And because the ball is not moving before it is fired so viy = 0 and the first term disappears leaving:

y = ½ gt2

But we want to solve for t, so we multiply both sides by 2 and divide both by g to get:

2 y/g = t2

2 x 100m/9.8 m/s2 = t2

t = 4.52 s

Whether the cannonball is dropped off the cliff, or fired horizontally, it takes 4.52 seconds to fall 100 m.

Now that we know the time we can use Eqn. 1.11 to calculate the velocity:

x = vixt + ½ axt2

This time ax is zero so the second term of the equation disappears:

x = vixt

and we rearrange to solve for velocity:

x/t = vix = 400 m/4.52 s = 88.5 m/s