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Projectile Motion

Hi there, in this post I will briefly try to explain what projectiles are. But… you might be thinking

Well, we can use projectiles in a lot of stuff like for example, shooting a basketball, kicking a football with an angle to achieve maximum possible horizontal distance(spoiler alert: the angle should be 45 degrees). The possibilities are endless, Well they are not endless but you get the point right?

At the end of the blog you will be able to deal with problems related to projectile motion AKA 2-dimensional motion.

Prerequisites

For understanding projectiles you need to know equations of motion(kinematic equations) and you should know some basic 1 dimensional motion. Other than you should know some basic algebra(to manipulate the equations) and some basic trigonometry.

What is a projectile?

Projectiles are basically a fancy way of saying that an object (say, a ball) travels in both vertical and horizontal direction simultaneously with the passage of time . For example when you play catch you throw the ball to your friend who is standing in front of you, the way the ball curves or moves in the air i.e. the path it takes to reach your friend forms an inverted U shape. This is called a projectile.

(Note: and if air resistance is negligible then the path of the ball/object will be a perfectly symmetrical if cut from the center as shown above).

But it doesn’t have to be perfectly symmetrical, for example when an object is thrown on an inclined mountain from the level ground it isn’t perfectly symmetrical. Example of projectile motion in angrybirds (ik no one plays that but it makes a good example)

Here is the formal definition in case you want to know: “A projectile can be defined as an object which is launched into air and the only force acting on it is gravity.”

Now lets find out which of these can be characterized as projectiles-

Exercise:

Question 1: Identify which of the following is/are projectiles and which of them are not?

a) A cargo plane moving with high speed ejects a capsule, the path traced by the capsule when it lands on the ground.( assume the capsule is not attached to a parachute). Hint: Think with respect to inertia

b) A swimmer jumps from a diving board from a height in a swimming competition by jumping a long distance into the pool when the whistle is blown, the path covered by the swimmer.

c) A bullet is fired vertically into the air from the surface of mars and reaches a maximum height of 10 units(meters./ft). The path covered by the bullet.

d) A tow truck pulls a car across the road, the path covered by the car.

e) A players hits a baseball, the path covered by the ball.

Bullet dropped vs fired

If I drop a ball and at the same time from the same height shoot a bullet horizontally(consider no air resistance in the vertical direction), BOTH OF THEM WILL HIT THE GROUND AT THE SAME TIME!!!!. I know this sounds absurd but this is true because of the fact that when you shoot the bullet horizontally it has no vertical speed(or velocity) and we also know that gravity has the same effect on all the objects irrespective of their mass(neglecting air resistance watch this if you are not convinced), hence we can say that the time taken by an object dropped from a height and a bullet shot horizontally takes the same amount of time to reach the ground. If you are not convinced then you can watch this experiment that beautifully demonstrates this point.

Sign convention and coordinate systems

Just like the sign convention in 1 dimensional motion, 2 dimensional motion also has a sign convention. I will be taking the vertically upwards direction as positive, vertically downwards as negative and horizontally rightwards direction as positive, horizontally leftwards direction as negative. But you are free to chose your own sign convention.

Given a point say P from the origin(origin basically means the point where we start measuring stuff from, in the above picture the point of intersection of x and y axis is the origin)

The coordinate of any point is usually written as (x, y) where x represents the horizontal position and y represents the vertical position. For example if a point is moved 3 units horizontally and 5 units vertically it would be written as (3,5). This kind of coordinate system is know as the Cartesian Coordinate system.

Finally projectiles

Projectiles are basically a combination of horizontal and vertical motion. This is a really useful fact, because now it just turns a projectile into two 1D motions which are independent of each other, the only thing that really binds them together is the fact that the time taken in both of the 1D motions is constant(Time of flight is constant). It doesn’t matter how fast the horizontal velocity is, if the time of flight (which we can get by solving the 1D motion for vertical direction) is say, 10 sec the object will hit the ground and come at rest i.e. the horizontal velocity will become zero.

You can try Example one given below yourself before reading solution. There is nothing new that you need to know!!!

Example-1: A football player kicks a ball with a speed of 20m/s at an angle 30 degrees with the x direction. Find the horizontal distance(Range) of the ball, Time of flight and maximum height attained by the ball during it’s flight.

Hint 1: First we need to break the motion of the projectile into two directions usually we break it into horizontal and vertical directions. We can also use axes for this purpose, i.e. x-axis for horizontal direction and y-axis for vertical direction.

Hint 2: The next step is to use “equations of motion” for x and y axes separately and find the missing values.

Solution: The question is asking us to find 3 things the Range(R), Time of flight(T) and the maximum height(H) attained by the projectile.

Range(R) is basically the horizontal distance(displacement) covered by an object, in this case the range is the distance between the launching point of the ball and the landing spot of the ball.

Time of flight(T) is exactly what it sounds like, it is the total time taken for the entire journey of the object.

Maximum Height(H) attained is very intuitive, it is shown below for this example.

In the above picture Hmax represents the Maximum Height(H), and the horizontal distance is the line joining the points P and Q.

IMPORTANT NOTE: We can also observe that in this given example the vertical displacement is zero. This is because if your see the coordinates of P and Q they are (0,0) and (R,0) respectively now to find the displacement we can use this formula final position – initial position i.e. we subtract x and y components individually so we get (R-0, 0-0) i.e. (R,0) so as you can see the the y coordinate of the displacement if zero we can say that the vertical component is zero(again, because y-axis represents the vertical direction).

We can apply second equation of motion in the y direction. To get the time of flight, T.

Please keep in mind the sign convention. for example in the above picture we have written -g instead of g because gravitational acceleration is in downwards direction.

Note that we first solved for y direction coz only one variable i.e. time T was missing

from the above picture we got the Time of flight T = 2 seconds. Now we can use the second equation of motion again but this time in the x-direction.

Hence we get the horizontal distance i.e. Range = 20√3 meters.

Finding the maximum height is not as straightforward as the above two.

We first need to argue that if we cut the curve along the maximum height vertically it splits into 2 symmetrical shapes. (This is because there is no acceleration in x-direction). So the time required to reach the maximum height is half of the time of flight T i.e. the time taken to reach the maximum height say, Thmax is T/2 i.e. Thmax = 1 second.

Yay! This question is finally complete.

Now let’s practice some more questions.

Question 1: A ball is thrown upwards. It returns to the ground describing a parabolic path. Which of the following remains constant? (air resistance is negligible)

Hint: Think in terms of the direction in which acceleration due to gravity is acting

a) Speed of the ball

b) Kinetic energy of the ball

c) Vertical component of velocity

d) Horizontal component of velocity

Question 2: During projectile motion, acceleration of a particle at the highest point of its trajectory(path) is: (consider air resistance to be negligible)

a) 9.8 m/s2 in downward direction

b) zero

c) 100ln(cos(iπ)) mole/m3

d) dependent upon the initial velocity at which the object is fired.

Question 3: The velocity of projection of a projectile is 6 m/s is rightwards and 8 m/s is upwards. The horizontal range of the projectile is:

a) 4.9 meters

b) 9.6 meters

c) 19.6 pounds

d) 30 feet 11 inches

Solutions to the questions

Question 1 solution: Since the air resistance is negligible and gravitational acceleration only acts in the vertical direction we can say that acceleration in the horizontal direction is zero. Hence the horizontal component of velocity will stay constant. so d option is correct

Question 2 solution: Acceleration in the horizontal direction is zero and the acceleration throughout the motion in the vertical direction is ‘g’ i.e. 9.8 m/s2 so a option is correct

Question 3: So basically you have to break this projectile into two 1D motions(along vertical and horizontal direction). Then you can solve the 1D motion in vertical direction and the1D motion in the horizontal direction(the order doesn’t matter). Then you just have to manipulate these 2 equation to find the range, which is option b) 9.6 meters.