I am going to break down this topic for you to understand the different aspect of projectile. Let's dive straight into it.
We are done with that phase. Let's go to the next phase of projectile.
What is a projectile?
A projectile is any object thrown into the air and moves freely on its own under the influence of gravity and air resistance.
What are Projectiles used in warfare?
A fired bullet, a shot arrow, a thrown bomb etc
What are Projectiles used in sports?
A thrown javelin, a ball kicked into the air, a thrown discuss etc
Let's take a look at the diagram below
When ever a question is given that relates to the diagram above we use the following formulae.
To find Height(H)
To find horizontal distance or Range.
There is another method using these two formulae.
Let's take some questions to understand this principle
- A particle is projected horizontally at 10m/s from a height of 45m. Calculate the horizontal distance covered by the particle before hitting the ground (g= 10m/s²)
Horizontal distance also means range
- A stone projected horizontally from the top of a tower with a speed of 4m/s lands on the level ground at a horizontal distance 25m from the foot of the tower. Calculate the height of the tower.( g= 10m/s²)
Let's take a look at this diagram
U represent initial velocity which can be resolved into two ways: horizontal component, Ux = Ucos𑁜 and vertical component, Uy = Usin𑁜
Note: At the maximum height, the vertical component of the initial velocity, Uy = 0
Terms to know
- Time of flight: This is the time it takes the projectile to get to the same level from which it was projected. Before we can calculate the time of flight, we need to learn how to calculate the time it takes the projectile to get to the maximum height. The formula is this
Where 't' - time to get to maximum height
Time of flight, T = 2t, which is
- Range: This is the horizontal distance from where it was projected to the point where it hits the projection plane. The formula for range is
To get the maximum range, the angle of projection, 𑁜 must be 45°
- Maximum height: This is the highest height reached as measured from the horizontal projection plane. The formula is
Let's us look at some examples to understand this better
Examples
- A body is projected at an angle of 30° to the horizontal with a velocity of 150m/s. Calculate the time it takes to reach the greatest height. (g= 10m/s²)
Solution
- The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50m/s. If the acceleration of free fall due to gravity is 10m/s². Determine its initial speed.
Solution
- An object is projected from a height of 80m above the ground with a velocity of 40m/s at an angle of 30° to the horizontal. The total time of flight is ?(g= 10m/s²)
Solution
First, calculate the time it takes the object to move from A to B, after which calculate the time it moves from B to C, them sum them up
- A bullet is fired at an angle of 60° to the vertical with a speed of 50m/s. Determine its maximum height reached. [Take g= 10m/s²]
Solution
We always use angle towards the horizontal, so we we going to subtract 60 from 90 to get angle towards the horizontal.
- A projectile is fired with a velocity of 20m/s at an angle of 40° to the horizontal. Determine the components of the velocity of the projectile at its maximum height.
Solution
- At what angle must a bullet be fired if its range must be equal to its maximum height?
Solution
- A shooter wants to fire a bullet in such a way that its horizontal range is equal to three times its maximum height. At what angle should he fire the bullet to achieve this?
Solution
Exercises
- A tennis ball projected at an angle attains a range R=78m. if the velocity imparted to the ball by the racket is 30m/s. calculate 𑁜
- A particle is projected horizontally at 15m/s from a height of 20m. Calculate the horizontal distance covered by the particle when it hits the level ground.
- A body is projected horizontally with a velocity of 15m/s from a cliff of height 180m.What is the range of the body ?(g= 10m/s)