# A cannon shoots a projectile at 24 degrees from the horizontal. It lands on level ground 3000m down range. a) Find the initial velocity b) Calculate its maximum height?

##### 2 Answers

#### Explanation:

Treat the vertical and horizontal components separately and then eliminate the time of flight.

The horizontal component of the velocity is constant so we can write:

For the vertical component we can use the general equation:

This becomes:

Note we divide the total time of flight by 2 because the time to reach maximum height (what we want) is the total time of flight / 2.

Now we can substitute the value of

We can simplify this using the trig identity:

This gives

However, we need to get **v** :

Putting in the numbers:

We can use:

This becomes:

a) The initial velocity is 199 m/s; b) The maximum height is 334 m.

#### Explanation:

The general path of a projectile is shown below.

We can derive an equation for the initial velocity

For motion in the **horizontal direction**,

For motion in the **vertical direction**,

At **maximum height**,

∴

The time to reach the top of the parabola is

It will take the same time for the cannonball to fall, so the total time in the air is

Knowing the time allows us to find the **range** of the cannon:

Rearranging gives

**b) Maximum height**

The distance traveled by a falling object is

Therefore, the maximum height is