Projectile motion is the motion of an object who's path is affected by the force of gravity. We are all affected by gravity, but it profoundly alters the motion of objects that are thrown or shot upward. The arching of a thrown ball is caused by gravity, as well as its falling motion in general.
Gravity is a force that acts on objects, it makes objects accelerate "downward". While we do not need to know about forces to analyze projectile motion we do need to know a very important detail: gravity causes objects to accelerate downward at roughly 9.8 m/s² 9.8 m/s² is the generally accepted amount of acceleration that happens, in some areas of Earth it is more or less but we will use 9.8 for our calculations.
Lets think of an object thrown straight up.
It is known that if an object is thrown up with a speed of V, it will land at the same spot with speed V, only the velocity will be downward. Lets investigate this interesting phenomenon.
We throw a rock from our hand. When the rock leaves our hand its travelling at
3m/s [up], an acceleration of 9.8m/s² [down] is acting on it and the height
at which it was released is considered to be zero. We know that at some point the
rock will come to rest, and the speed will be zero. Instead of using vectors we
will say that upward motion is positive and downward motion is negative.
Vi = 3m/s
Vf = 0m/s rock stopped in mid-air
a = -9.8m/s²
d = ?m
t = irrelevant
Lets arrange our uniform acceleration formulae to remove the time variable and find out the height at which the rock will stand still.
2da = Vf² - Vi²
d = ( Vf² - Vi² ) / 2a
d = ( 0 - s )² / 2( -9.8 );
d = -9 / -19.6
d = 0.46
The rock stops in mid-air 46cm above your palm.
Now, with the information that we know, lets find out at what speed it will drop back to your hand:
Vi = 0m/s wer'e starting up suspended in the air
Vf = ?m/s
a = -9.8m/s²
d = 0.46m
t = irrelevant
We can actually use the same formula, how convenient!
2da = Vf² - Vi²
Vf² = 2da + Vi²
Vf² = 2( 0.46 )( -9.8 ) + 0²
Vf² = 2( 0.46 )( -9.8 )
Vf² = -9.016
Vf = -3.0 m
The speed at which the rock hits your hand is indeed equal to that with which it was thrown and opposite in direction. A keen observer may have noticed that at the last line of the calculation the square root of -9.016 was taken. While this is a mathematical impossibility and will yield a complex number ( one that doesn't actually exist ) it is perfectly fine for a physicist to do, since the negative sign serves only to tell us the direction of motion - not its maginitude.
We now know that whenever an object is in the air it is being accelerated at 9.8m/s² downward, but is there a corresponding horizontal acceleration? The only acceleration acting horizontally is deceleration due to air friction, this is often ignored and it is considered that there is no horizontal acceleration whatsoever. The only horizontal acceleration on an object happens when your hand acclerates it from rest, its interesting to note that once an object leaves your hand it will not slow down horizontally.
Lets do an example where an object is thrown horizontally, with no vertical force acting on it, to examine that type of situation.
John throws a frisby off the top of a hill 20m high, he throws it with a speed of 8.6m/s, at what speed will the frisby hit the ground below the hill? John's hand is 1.5m above the surface of the hill.
At first glance it seems that we are given an insufficient amount of data but by combining our knowledge of uniform acceleration and projectile motion we will be able to do solve this question.
In the diagram above you can see John performing the aforementioned actions. The frisby starts off 20 + 1.5 metres above the ground ( hill + john's hand's height ), it has a velocity to the right and it has a velocity down. The horizontal velocity is given to us, and the frisby will maintain it for the whole duration of the motion. The downward velocity is not constant, it accelerates constantly with a magnitude of 9.8m/s² [down] and we will have to use the uniform acceleration formulae to find out the downward speed of the frisby upon impact. The blue vector is the resultant velocity, the magnitude of which we need, and is found by adding the two black vectors together. The green line is the frisby's physical path.
Lets list the data that we know, along with the direction ( horizontal/vertical ) it is associated with.
|Horizontal (x)||Vertical (y)|
Vix = 8.6m/s
Vfx = 8.6m/s
tx = ?s
dx = 20+1.5 = 21.5m
ax = 0
Viy = 0m/s
Vfy = ?m/s
ay = 9.8m/s²
ty = tx
dy = ?m
Lets consider the right direction and the downward direction to be positive. The vertical and horizontal components of motion can be related to each other by the time. Since the time of the motion is the same in both directions we can find it using the horizontal data we know.
d = 1/2 ( Vfx + Vix ) × t
t = 2d / ( Vx + Vx )
t = 2( 21.5 ) / 2( 8.6 )
t = 21.5 / 8.6
t = 2.5s
Now that we know that the whole motion took 2.5 seconds we can use uniform acceleration formulae to find out Vfy.
a = ( Vfy - Viy ) / t
Vfy = ta + Viy
Vfy = (2.5)(9.8) + 0
Vfy = 24.5m/s
Now that we know that the components of the final velocity are 8.6m/s [right] and 24.5m/s [down] we can use the pythagorean theorem to find the impact speed of the frisby saucer:
Vimpact² = 8.6² + 24.5²
Vimpact² = 73.96 + 600.25
Vimpact² = 674.21
Vimpact = 25.96 m/s
We conclude that when the frisby lands it has a speed of 26m/s . That's quite a gain from the starting speed of 8.6m/s, all of that extra speed is due to gravity.
Lets convert the units of the impact speed into kilometres per
hour to give us an idea of the frisby's fun level.
( 26m/s × 3600 ) / 1000 = 93.6 km/h This frisby is as fast as a car, and will probably create a crater in the ground.
Situations where an object is projected upward at an angle are much more common than those where an object is projected straight up or completely horizontally. Thus they are more useful to us.
How do we relate an object's takeoff velocity to its impact velocity? We know that the takeoff velocity and impact velocity have two components, the horizontal and vertical. We know that whenever an object is projected its horizontal velocity remains constant, while vertically it is accelerated at 9.8m/s² [down].
The object starts its motion with a certain horizontal velocity and ends the motion with the same horizontal velocity. About the vertical velocity we know that when an object is thrown up it lands with the same speed as it left, therefore the only change in the components of the velocities ( takeoff and landing ) is that the vertical velocity is now in the downward direction. Since the magnitudes of the components are all the same, the speed of takeoff and landing will be identical. The only difference between the velocities is the direction - the impact velocity will have the same angle as the takeoff, only below the horizontal.
The above means that if takeoff velocity is 5m/s [Left 66² Up] the impact velocity will be 5m/s [Left 66² Down].
This example was very popular in the days of yore. A buccaneer is attacking a gold bearing ship. The buccaneer's cannons can shoot projectiles at a speed of 70.0m/s, the cannons are tilted at 30° up. How close does the buccaneer's ship have to get to the transport ship for the cannon ball to hit it?
i = 70m/s
a = -9.80m/s²
ß = 30°
The height of the boats above the water is negligible.
Let the up direction be positive.
We need to calculate how far, horizontally, the projectile will fly before hitting the water.
Vx = 70 × Cos 30° = 60.6m/s
Viy = 70 × Sin 30° = 35.0m/s
Vfy = - Viy
t = ( Vfy - Viy ) / a Time of vertical flight
t = ( -35 - 35 ) / -9.8
t = 7.14 s Is the same as horizontal flight
Lets see how far from the pirate ship the cannon ball will fall.
d = Vx × t
d = 60.6 × 7.14
d = 432.68
The buccaneer has to bring his ship to within 432 metres of the victim ship.
Projectile motion is easy to understand - objects don't accelerate horizontally but they decelerate vertically. Projectile motion is in action every time an object is thrown, dropped, ejected or shot.
To solve projectile motion problems you need to relate the motion's horizontal component to its vertical component. Most times the time will relate the components together, because the object moves horizontally only as long as it moves vertically.
Gravity decelerates an object at a constant rate, and an object will fall past its takeoff point with the same speed as it took off with. Next Section: Relative Velocity