This animation shows the parabolic path of a ball rolled off of something like a table and then allowed to fall freely. Click play to set it in motion. The projectile is the magenta ball; the green balls represent its horizontal and vertical velocities.

Notice that the horizontal velocity is **constant**, while the downward motion is **accelerated**. Run the animation a few times looking at different aspects of the motion. Look at both the forward progress (top) and the downward progress (left) of the ball.

The essence of how we treat projectile motion, the motion of a launched object after no more launching forces are working on it, is in this figure: The horizontal velocity (ignoring friction) is constant, and the vertical acceleration is just that of a freely-falling object.

A **projectile** is any object set free of any forces except for gravity and friction. A projectile can be a thrown ball, a bullet or a springboard diver ...

Except for air resistance, the forward velocity of any projectile is constant and is equal to the initial velocity when it was released. The vertical velocity changes by the acceleration of gravity.

Nature doesn't care that a projectile is moving forward. It's downward acceleration is just the acceleration of gravity, exactly as though it was dropped or thrown straight upward.

Consider the forces at work on a ball while it's rolling along the table. We assume that a force was or is being applied that created a constant velocity **v _{x}**. The force of gravity,

Now as the ball rolls off the table, the normal force vanishes and the force of gravity pulls it downward. Unbalanced forces produce acceleration, so the ball will accelerate downward at 9.8 m/s^{2}. There is no force acting horizontally (except for friction – air resistance), so the forward velocity will remain **v _{x}**

As the ball falls, two things are going on **independently**. First, the ball is traveling forward at a constant velocity. You can see that by the even spacing of the tick marks along the horizontal axis in the figure below.

As time unfolds, the downward velocity of the ball increases at the rate of 9.8 m/s^{2}, creating the uneven spacing in the vertical location of the ball (ticks on vertical axis).

That's the essence of projectile motion, no matter how complicated the scenario: Nature doesn't care about whether a projectile is moving horizontally. It's still going to be acted on by the force of gravity just as though it were dropped or thrown straight upward.

A ball rolls off the top of a 1.5 m tall table with a horizontal velocity of 2.0 ms^{-1}. Calculate how far the ball will travel from the table before it hits the floor.

**Solution**

When the ball rolls off the edge of the table, it will continue moving forward at 2.0 m/s until it hits the floor. So we just need to find out how long it will be in the air before it hits.

For that, we'll use the freefall formula, rearranged to give us the time. What's crucial here is to realize that, when it comes to the vertical dimension, the ball is acting just like a dropped ball. Nature doesn't care that it's also moving forward at the same time.

$$d = \frac{1}{2} gt^2 \; \longrightarrow \; t = \sqrt{\frac{2d}{g}}$$

Plugging in the height of the table and the acceleration of gravity, we get the time in seconds:

$$ \require{cancel} t = \sqrt{\frac{2(1.5 \cancel{m})}{9.8 \cancel{m}\cdot s^{-2}}} = 0.553 \; s$$

So the ball will be in the air (and thus able to move forward) for a little more than half a second. Now we just rearrange the speed equation,

$$s = \frac{d}{t} \; \longrightarrow \; d = st$$

and plug in the speed and time to get the distance traveled:

$$ \begin{align} d &= 2.0 \frac{m}{\cancel{s}} (0.553 \cancel{s}) \\[5pt] &= \bf 1.11 \; \text{m} \end{align}$$

The ball will travel about 1.1 meters before hitting the ground.

Projectile motion problems always have this kind of two-step structure (at least).

A baseball is hit so that it takes off at a 45˚ angle with respect to the ground and with an initial velocity of 40 m/s. How far would the ball travel if there were no fence or wall to hit?

**Solution**

We'll need to break that initial velocity vector into its vertical and horizontal components to get the vertical velocity, vy and the horizontal velocity **v _{x}**. To do that we can use the 45-45-90 triangle, which you should memorize.

The length of a side of the 45-45-90 triangle is $\sqrt{2}/2$ times the length of the hypotenuse, so our hitter diagram looks like this:

Now to find the total time in the air, we use the definition of acceleration to find the time it takes the ball to get to the top of its arc, the acceleration in this case being $g = 9.8 m/s^2.$ We can rearrange that definition to find the elapsed time.

$$g = \frac{\Delta v}{\Delta t} \; \longrightarrow \; \Delta t = \frac{v_f - v_i}{g}$$

We are able to calculate the time because the velocity at the top of the arc, **v _{f}**, is zero. The one-way time is:

$$\Delta t = \frac{0 - 20\sqrt{2}}{-9.8} = 2.886 \; s$$

The round trip time – up and down – is twice that, 5.772 s. Now it's just a matter of calculating the distance traveled during that time, given that the horizontal velocity is $20 \sqrt{2} \, m/s.$

$$ \begin{align} &= 20 \sqrt{2} (5.772) \\[5pt] &= 163 \, m \; \; \text{ or 535 ft.} \end{align}$$

This is a pretty typical projectile motion problem that you should know how to solve.

Remember that this distance is really an upper limit to the distance that the ball could travel. In reality, air resistance would slow its travel a bit and the distance would decrease.

A certain cannon has a muzzle velocity of 90 m/s. Calculate the distance a cannonball would travel if launched at angles of 40˚, 45˚ and 50˚ with respect to flat ground.

**Solution**

The question we're really asking here is, which angle is the best for achieving the largest distance of the shot? We'll guess that it's 45˚ and test an angle on either side, 40˚ and 50˚.

This problem will require a little trigonometry because not all of those angles belong to convenient memorized triangles. The vertical (**v _{y}**) and horizontal (

Let's calculate those vector components of the velocity for each angle using the sines and cosines:

Now we need to calculate the total time in the air using the definition of acceleration and the fact that the vertical velocity at the top of the arc is zero. The factor of 2 in this equation accounts for the round-trip up and back down.

Here are the results for each angle:

Now to calculate the distances using the horizontal velocities and the time interval,

$$d = v_x \cdot \Delta t$$

Finally, here are the distances:

Indeed, the 45˚ angle yields the greatest distance. This is always true for a given initial velocity, all other conditions being equal.

Let's look at another cannon problem. A cannon (muzzle velocity 110 m/s) located at the edge of a 50 m cliff is shot upward at a 55˚ angle. How far from the base of the cliff (assume that the cliff wall is vertical) will the cannonball land? Assume that the cannon muzzle is 1.5m above the flat ground on top of the cliff.

**Solution**

In this problem, we'll be a little more precise and take the height of the muzzle of the cannon into account, too.

The first step is to calculate vertical and horizontal components of the initial velocity, **v _{y}** and

$$ \begin{align} v_x &= 110 \cdot cos(55˚) = 63.09 \; \frac{m}{s} \\[5pt] v_y &= 110 \cdot sin(55˚) = 90.11 \; \frac{m}{s} \end{align}$$

Now to calculate the distance and time to the top of the arc. The total time will have to be calculated in two steps because the up and down distances are different.

$$\Delta t_{up} = \frac{0 - v_y}{g} = 9.195 \; g$$

We'll need the upward distance in order to calculate the time to the bottom of the cliff.

$$ \begin{align} d_{up} = v \cdot \Delta t_{up} &= \frac{90.11 \, \frac{m}{\cancel{s}} (9.195 \cancel{s})}{2} \\[5pt] &= 414.3 \; m \end{align}$$

The velocity in that last step is the average velocity of the climb, the average of 90.11 m/s and 0.0 m/s. Now the time it takes for the projectile to fall is calculated from a rearranged freefall equation and by adding 50 m + 1.5 m to the upward distance:

$$ \begin{align} \Delta t_{down} &= \sqrt{\frac{2d}{g}} \\[5pt] &= \sqrt{\frac{2(414.3 + 51.5)}{g}} \\[5pt] &= 9.749 \, s \end{align}$$

Now the total time that the projectile spends moving forward (before it hits the ground) is

$$ \begin{align} \Delta t &= 9.749 + 9.195 \\[5pt] &= 18.944 \; s \end{align}$$

Finally, as always, we use the horizontal velocity and the total time to calculate the distance traveled:

$$ \begin{align} d = v_x \cdot \Delta t &= 63.09 \frac{m}{\cancel{s}} \cdot 19.944 \cancel{s} \\[5pt] &= 1258 \; m \end{align}$$

That's pretty far. The cliff gives us an advantage. Can you think of a simple change that would give us even more distance?

Step | Procedure |
---|---|

1. | If necessary, resolve the initial velocity vector into horizontal (v) and vertical (_{x}v) components_{y} |

2. | Use v to calculate the total time in the air. Take extra care if the projectile doesn't originate and land at the same height._{y} |

3. | With the time, calculate how far the projectile would move horizontally at a constant velocity of v_{x} |

1. | A ball is launched straight upward with an initial velocity of 3.0 m/s. The point where the ball exits from the mechanical launcher is 0.25 m from the ground. How high above the ground will the ball go? | |

2. | If a ball shot vertically rises to a height of 2.35 meters, what was its initial velocity? | |

3. | A ball rolls toward the edge of a 37 m tall vertical cliff at a velocity of 5 m/s. Calculate how far the ball will be from the cliff when it hits the ground. | |

4. | A suitcase was dropped from a plane traveling at 300 m/s at an altitude of 35,000 ft. Calculate the horizontal distance, between the point where the suitcase was dropped and the point where it landed. Assume that there is no air resistance. (1m = 3.28 ft.) | |

5. | A rifle with a muzzle velocity of 829 m/s is fired at a 32˚ angle with respect to the ground. Ignoring air resistance, how far away would the bullet land. Assume the bullet is fired from ground level. Does your answer make sense? | |

6. | Consider the case of a shot putter throwing a 9 lb. shot put in a competition. To achieve optimum distance, the thrower hurls the steel ball at a 45˚ upward angle. He can generate enough acceleration of the shot to achieve an initial velocity of 14.1 m/s. How far will the shot put fly? Assume that the release point is 8 ft. off the ground and that the ball lands on level ground. The world record in the men's shot put event is 23.12 m |

Here are four examples of how to solve projectile-motion problems.

A ball rolls off a table with a given initial velocity. How far from the edge of the table will it hit the floor?

Minutes of your life: 0:00

A cannon is fired on level ground at a given muzzle velocity (initial velocity) and a 30˚ angle. How far will the cannonball travel before hitting the ground?

Minutes of your life: 0:00

A cannon is fired from the top of a high cliff with a given muzzle velocity and angle with respect to the horizontal. Calculate the horizontal distance that the ball will travel.

Minutes of your life: 0:00

A cannon is fired with a muzzle velocity of 30 m/s at a 35˚ angle toward a 30 m high plateau 120 m away. Will the ball clear the top of the plateau?

Minutes of your life: 0:00

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