projectile motion

What is projectile motion

Projectiles are objects thrown into the air, and in this chapter, we will study how they travel.

Projects can either be thrown in many ways. You can throw an object horiontally above a cliff, at an angle, or even vertically. We will study all of these scenarios.

Horizontal motion

Horizontal motion is quite simple. In this course, and in your high school course, we will ignore factors that may affect the horizontal motion of a projectile in the air, such as friction, and treat it as if there are no external factors affecting it.

A a result, the horizontal speed of the projectile remains constant throughout the ball's motion. So, v⃗xi=v⃗xf \vec {v}_{x_i} = \vec {v}_{x_f} .


Therefore, the only equation we need that describes the horizontal motion of a ball is:

v⃗x=d⃗xt \vec {v}_x = \frac { \vec {d}_x } { t }

Where v⃗x \vec {v}_x is the horizontal velocity of the ball, d⃗x \vec {d}_x is the horizontal displacement of the ball, and tt is time.

Vertical motion

When considering the vertical motion of the projectile, the only considerable force being applied onto it would be gravity. It causes an acceleration of approximately 9.81ms2 9.81 \frac {m} {s^2} downwards.

Again, we will not be considering the force of friction caused by air.


Here are the equations that you will need to use in order to solve projectile motion problems in regards to vertical motion:

g⃗=v⃗yf − v⃗yitd⃗y=(v⃗yf−v⃗yi2)td⃗y=v⃗yit+12g⃗tv⃗yf2=v⃗yf2+2g⃗d⃗y \vec {g} = \frac { \vec {v}_{y_f} \ - \ \vec {v}_{y_i} } {t} \newline \vec {d}_{y} = ( \frac { \vec {v}_{y_f} - \vec {v}_{y_i} } {2} )t \newline \vec {d}_{y} = \vec {v}_{y_i} t + \frac {1} {2} \vec {g} t \newline \vec {v}_{y_f}^2 = \vec {v}_{y_f}^2 + 2 \vec {g} \vec {d}_{y}

Where v⃗yi \vec {v}_{y_i} and v⃗yf \vec {v}_{y_f} are the initial and final vertical velocities, d⃗y \vec {d}_{y} is the vertical displacement, g⃗ \vec {g} is the acceleration due to gravity (9.81ms2 9.81 \frac {m} {s^2} ) and tt is time.

Visualization

Here is an animation of what those vectors look like on a projectile being thrown on a level field:

Clearly, you can see that the horizontal velocity does not change, however, the vertical velocity is at a maximum at when the projectile launches, and when it lands. Something else to note, is that the vertical velocity is equal to 00 when the projectile hits its maximum height.

Tips for dealing with projectile motion

  • If the projectile is thrown on a level field, the vertical velocity v⃗y \vec {v}_y is at a maximum at launch (t=0 t = 0 ), as well as when it lands (t=tf t = t_f ).

  • The vertical velocity v⃗y \vec {v}_y is always equal to 00 when at its maximum achievable height (d⃗ymax \vec {d}_{y_{\text {max} } } )

  • If the projectile is thrown on a level field, the maximum height (d⃗ymax \vec {d}_{y_{\text {max} } } ) always occurs at the midpoint between the launch and landing of the projectile.

Projectile motion on a level field

Questions like these are quite easy because you can utilize many of the tips previously mentioned.

The image below demonstrates what the motion of the projectile looks like:

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Example

A ball is thrown at a velocity of 7.50ms 7.50 \frac {m} {s} into the air 45.0∘ 45.0^{\circ} . At what time is the ball at its highest point?

Answer

We know that the vertical velocity of the ball will be equal to 0ms0 \frac {m} {s} when it is at its highest point. Knowing this, g⃗=v⃗yf − v⃗yit \vec {g} = \frac { \vec {v}_{y_f} \ - \ \vec {v}_{y_i} } {t} would be the best equation to use:

    g⃗=v⃗yf − v⃗yit \quad \quad \ \ \ \, \vec {g} = \frac { \vec {v}_{y_f} \ - \ \vec {v}_{y_i} } {t}

−9.81ms2=(0) − (7.50m)(cos⁡45.0∘)t -9.81 \frac {m} {s^2} = \frac { (0) \ - \ (7.50m)(\cos 45.0^{\circ}) } {t}

     t=−(7.50m)(cos⁡45.0∘)−9.81ms2 \quad \quad \ \ \ \, \, t = \frac { -(7.50m)(\cos 45.0^{\circ}) } {-9.81 \frac {m} {s^2}}

     t=0.54s \quad \quad \ \ \ \, \, t = 0.54s

∴ \therefore The ball reaches its maximum height at t=0.54s t = 0.54s .

Projectile motion on a cliff

Projectiles thrown off of a cliff can be somewhat harder to solve, since it may not be as obvious which equation you need to use.

It may help to list the equations you already have in a "given" list.

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In the figure above, the ball is thrown slightly away from the edge of the cliff. Unless otherwise mentioned on some tougher questions, assume it is thrown at the edge itself. The image was there just to show the trajectory of the ball.

Example

A rock is thrown horizontally at the edge of a cliff 30.0m 30.0m tall at a velocity of 15.0ms 15.0 \frac {m} {s} . How far from the base of the cliff does the ball land?

Answer

Let's start with our given:

  • d⃗y=−30.0m \vec {d}_y = -30.0m

  • v⃗xi=15.0ms=v⃗i \vec {v}_{x_i} = 15.0 \frac {m} {s} = \vec {v}_i

  • g⃗=−9.81ms2 \vec {g} = -9.81 \frac {m} {s^2}


Now, we need to determine at what time the rock hits the ground. Since the rock was thrown horizontally, it's maximum height occurs at t=0 t = 0 . This means that v⃗yi=0 \vec {v}_{y_i} = 0 .


So, the best equation to use would be d⃗y=v⃗it+12g⃗t2 \vec {d}_y = \vec {v}_i t + \frac {1} {2} \vec {g} t^2 :

   d⃗y=v⃗it+12g⃗t2 \quad \quad \, \, \, \vec {d}_y = \vec {v}_i t + \frac {1} {2} \vec {g} t^2

−30.0m=(0)t+12(−9.81ms2)t2 -30.0m = (0) t + \frac {1} {2} (-9.81 \frac {m} {s^2}) t^2

−60.0m=(−9.81ms2)t2 -60.0m = (-9.81 \frac {m} {s^2}) t^2

    t2=6.12s \quad \quad \ \, \, \, t^2 = 6.12s

     t=2.47s \quad \quad \ \ \ \, \, t = 2.47s

Notice how d⃗x \vec {d}_x is negative, and this is because the ball was displaced 30.0m 30.0m down after it was released. It is also the same sign as g⃗ \vec {g} , and as a result, you get a positive value for tt.


Now that we have the time at which the rock hits the ground, we can use the equation v⃗x=d⃗xt \vec {v}_x = \frac { \vec {d}_x } {t} to calculate how far from the base of the cliff it landed:

v⃗x=d⃗xt ⇒ d⃗x=v⃗xt=(15.0ms)(2.47s)=37.1m \vec {v}_x = \frac { \vec {d}_x } {t} \ \Rightarrow \ \vec {d}_x = \vec {v}_x t = (15.0 \frac {m} {s}) (2.47s) = 37.1m


∴ \therefore The rock fell 37.1m 37.1m from the base of the cliff.

Example

A ball is thrown at an angle of 30.0∘ 30.0^{\circ} off of a 9.00m 9.00m edge at a velocity of 5.00ms 5.00 \frac {m} {s} . How far does the ball land from the base of the edge.

Answer

Here's our given:

  • v⃗i=5.00ms \vec {v}_i = 5.00 \frac {m} {s}

  • d⃗y=−9.00m \vec {d}_y = -9.00m

  • g⃗=−9.81mss\vec {g} = -9.81 \frac {m} {s^s}


The best equation to use would be d⃗y=v⃗yit+12g⃗t2 \vec {d}_y = \vec {v}_{y_i} t + \frac {1} {2} \vec {g} t^2 , since we can easily determine v⃗yi \vec {v}_{y_i} by using the sin⁡ \sin function:

v⃗yi=v⃗isin⁡θ=(5.00ms)sin⁡(30.0∘)=2.5ms \vec {v}_{y_i} = \vec {v}_i \sin \theta = (5.00 \frac {m} {s}) \sin (30.0^{\circ}) = 2.5 \frac {m} {s}


d⃗y=v⃗yit+12g⃗t2 \vec {d}_y = \vec {v}_{y_i} t + \frac {1} {2} \vec {g} t^2

−9.00m=(2.5ms)t+12(−9.81ms2)t2 -9.00m = (2.5 \frac {m} {s}) t + \frac {1} {2} (-9.81 \frac {m} {s^2}) t^2

0=(−4.905ms2)t2+(2.5ms)t+9.00m 0 = (-4.905 \frac {m} {s^2}) t^2 + (2.5 \frac {m} {s}) t + 9.00m


By using the quadratic formula, we get:

t=−1.12s t = -1.12s or t=1.63s t = 1.63s

Obviously, we can't have negative time, so the ball lands at t=1.63s t = 1.63s .


Finally, we can determine the value of v⃗xi \vec {v}_{x_i} , then plug that into v⃗x=d⃗xt \vec {v}_x = \frac { \vec {d}_x } {t} :

v⃗xi=(5.00ms)cos⁡(30.0∘)=4.33ms \vec {v}_{x_i} = (5.00 \frac {m} {s}) \cos (30.0^{\circ}) = 4.33 \frac {m} {s}

v⃗x=d⃗xt ⇒ d⃗x=v⃗xt=(4.33ms)(1.63s)=7.06m \vec {v}_x = \frac { \vec {d}_x } {t} \ \Rightarrow \ \vec {d}_x = \vec {v}_x t = (4.33 \frac {m} {s}) (1.63s) = 7.06m


∴ \therefore The ball landed 7.06m 7.06m from the base of the edge.