Mechanics Tutorial 9  Projectiles
A projectile is any object that is thrown or released into a gravity field. There are three situations we will look at:
Throwing an object vertically into the air;
Throwing an object horizontally from a height;
Throwing an object at an angle.
The horizontal movement is totally independent of the vertical movement. That means that they do NOT affect each other.
The two movements are vector quantities, so they have a direction.
The velocities are at 90^{o} to each other.
There is a resultant velocity from the two independent velocities.
We can analyse the vertical movement using the equations of motion.
This topic needs you to be thinking very carefully.

General Concepts
We will call the horizontal motion the xcomponent, and the vertical motion the ycomponent.
At this level, we always ignore air resistance.
Let’s look at the horizontal motion:
v_{x} and s_{x} are the horizontal velocity and horizontal displacement respectively.
You can see that it’s very simple and it is affected in NO WAY by the vertical motion. If we want to know the range (the displacement of projectile when it lands), we need to know v_{x}_{ }and t.
Do
not introduce the vertical velocity into your treatment of the
horizontal velocity. 
Now let’s look at the vertical motion. Since we have an acceleration (g = 9.81 m s^{2}), we use the equations of motion that involve acceleration. We also need to be aware of the minus sign that tells us that the acceleration is vertically downwards.
The terms u_{y, } v_{y}, , and s_{y} stand for initial vertical velocity, final vertical velocity, and vertical displacement respectively.
The directions are important. If the vertical velocity is positive, it means that the object is moving upwards. This is also true for the displacement. Downwards is negative.
Whether the object is moving upwards or downwards, it is accelerating at 9.81 m s^{2}.
Some points from this diagram:
On the ground s_{y} is zero.
If the object carries on falling below the level from which it is thrown, s_{y} is negative.
At the very top of the projectile path v_{y} is zero, but the object is still accelerating at – 9.81 m s^{2}.
At all points v_{x} remains unchanged.
The horizontal velocity v_{x} is totally independent of v_{y}.
The path is a parabola.
Many students forget the independence of vertical and horizontal motions. They blindly put the horizontal velocity into the equations: ... ...usually into the u_{y} term. That will get zero marks in the exam. 
The example below includes what NOT to do.
Worked Example An aerobatic pilot is flying at a speed of 70 m s^{1} between airshows at a height of 500 m. He is hungry and eats a pork pie which he finds utterly unpleasant. He throws it out of the aeroplane. a) How long does the pie take to reach the ground? b) What is the distance between the point of release and the point at which it hits the ground? c) What is its speed when it hits the ground?

Answer

a) Use
500 m = 0 × t + ½ × 9.81 m s^{2} × t^{2}
t^{2} = (500 m × 2) ÷ 9.81 m s^{2}
t^{2} = 1000 m ÷ 9.81 m s^{2} = 102 s^{2}
t = Ö102 s^{2} = 10.1 s



Projectile problems often cause problems for students. They are not that difficult as long as you remember the problemsolving strategy. Let us look at three different situations.
Throwing An Object Vertically Into The Air
Consider a basketball player throwing a ball in the air. What goes up must come down.
Picture from a clip art collection
The ball has a downward force acting on it because of gravity. Therefore it will slow down at a rate of 9.81 m s^{2}. So we can say that the acceleration is 9.81 m s^{2}. When we tackle problems like this, we use the equations of motion. Click HERE if you need to review the equations of motion. We have to make sure that we get the signs right. We will make upwards positive and downwards negative.
The girl throws the ball at an upward velocity of 15 m s^{1}. How high will it go? 

How long will it take the ball to reach its maximum height? In both questions use g = 9.8 m s^{2} 
We can represent these motions graphically. It is important that you understand these graphs.
A displacement time graph looks like this:
The graph is a parabola because the ball is accelerating downwards. When it reaches the top, its velocity is 0, but its acceleration is still ()9.81 m s^{2}.
The velocity time graph looks like this:
This graph shows how the velocity not only changes, but its sign changes as well. This tells us that the direction changes as well. This stands to reason; if going up is positive, going down must be negative. Note that the gradient is constant, i.e. the acceleration is constant.
What would the speed time graph look like? 
The acceleration time graph looks like this.
It shows us that the acceleration is constant at 9.81 m s^{2}. The minus sign tells us that the acceleration is towards the ground.
If we throw an object
horizontally, there are two important things to consider:
The
horizontal velocity remains constant (ignoring air resistance)
The vertical
velocity increases at a rate of ()9.8 m/s^{2}.
If we throw one object and drop a second object at the same time, we see this:
The second object, thrown
horizontally, will hit the ground at the same time as the object that is simply
dropped.
Although the drawing is not to
scale, you can see how the horizontal velocity remains constant, while there is acceleration
downwards.
Look at the diagram
below. A pallet is dropped from a helicopter to the ground. We will
ignore the air resistance.
The path taken is NOT a straight line, because the velocity downwards is increasing at a constant rate of 9.8 m/s^{2}. It is a parabola. There are two components in this problem:
The horizontal velocity which remains constant.
The vertical velocity which changes, as the object is accelerating towards the ground. We use an equation of motion to analyse the motion.
The key point to remember is that the horizontal and the vertical motions are independent.
A common beartrap is to put the horizontal velocity into the vertical equation of motion. 
What is the horizontal velocity? 

Show that the vertical velocity about 44 m s^{1} towards the ground? Note that the horizontal velocity is ignored. Use g = 9.8 m s^{2} 

What is the resultant velocity of the pallet just before it hits the ground? 
Picture by Casito, Wikimedia Commons
Archery is a sport in which the participants subconsciously do calculations involving movement in two directions. Again the vertical and horizontal movement are independent.
Let us analyse the motion from the moment an arrow is released to the moment it hits the target. We want to find the range. For simplicity we will assume that the target is at the same height as the release point. We will also ignore air resistance. We will not worry about the signs.
1. On release, the arrow leaves at a velocity v m s^{1} and angle q. The horizontal velocity is v cos q m s^{1} . The horizontal velocity remains constant at v cos q m s^{1}.
2. The vertical velocity is v sin q m s^{1} initially.
3. To work out the time we need to use an equation of motion that has initial velocity, acceleration, and time.
The equation:
v = u + at
will fit the bill.
0 = v sin q + at
v sin q = at
Therefore
4. To get the range we need to multiply the horizontal velocity by the time taken in the air. Therefore:
range = v cos q × 2t
It is 2t because it takes t seconds for the arrow to go up to its maximum height and t seconds for it to come down again.
Worked example A large firework rocket leaves a launch tube at a velocity of 110 m s^{1} at an angle of 30 degrees. What is the range of the rocket? 
1. Work out the horizontal velocity. Horizontal velocity = v cos q = 110 m s^{1} x cos 30 = 110 m s^{1} x 0.866 = 95.3 m s^{1} 
2. Work out the initial vertical velocity: The initial vertical velocity = v sin q = 110 m s^{1} sin 30 = 110 m s^{1} x 0.5 = 55 m s^{1}. 
3. Now work out the time it takes to get to the maximum height: 0 = 55 m s^{1} + 9.81 m s^{2} x t (Þ 0  55 m s^{1} = 9.81 m s^{2} × t) t = 55 m s^{1} ÷ 9.81 m s^{2} = 5.61 s
Therefore the total time in the air = 2 x 5.61 s = 11.2 s 
4. Therefore the range = v cos q × 2t = 11.2 s × 95.3 m s^{1} = 1070 m. 
In the AS exam, they will not be over officious with signs, but make sure you explain each step. However it's good practice to get the signs right. In the A level exam, they will be expecting correct use of the signs for full marks.
A javelin thrower throws a javelin at a velocity of 25 m s^{1} at an angle of 40 degrees. What distance will he throw the javelin? Use g = 9.8 m s^{2} 
Finding the range without the time
Consider an arrow fired at an angle q at a velocity of v m s^{1}. The trajectory is like this:
We know that the horizontal velocity remains constant.
v_{x} = v cos q
We know that the vertical velocity changes. The vertical velocity at the start is:
vy = v sin q
So we can write:
uy = v sin q
The vertical velocity is 0 at the top of the trajectory. Work out time taken to get to top:
Therefore
Therefore we can write:
The range is s_{x}, and the time taken is 2t, because t is the time for the arrow to get to the top. What goes up must come down, and it takes exactly the same time to fall as it does to go up. Therefore we can write:
s_{x} = 2t × v_{x} = 2t × v cos q
Since:
We can substitute to get:
Note that g =  9.81 m s^{2}, so the minus signs cancel out. The time is always positive (you can't go back in time).
Worked Example A large firework rocket leaves a launch tube at a velocity of 110 m s^{1} at an angle of 30 degrees. What is the range of the rocket? 
Answer Use:
Put in the numbers:
The range is 1070 m to 3 s.f.
This is consistent with the answer in the previous worked example. 