Motion in a plane is also referred to as a motion in two dimensions. For example – Circular Motion, Projectile Motion, etc. For the analysis of such type of motion (i.e. Projectile Motion), the reference point will be made of an origin and the two coordinate axes X and Y. One of the most common examples of motion in a plane is Projectile motion.
Projectile refers to an object that is in flight after being thrown or projected. In a projectile motion, the only acceleration acting is in the vertical direction which is acceleration due to gravity (g). Equations of motion, therefore, can be applied separately in X-axis and Y-axis to find the unknown parameters.
Some examples of Projectile Motion are Football, A baseball, A cricket ball, or any other object. The projectile motion consists of two parts – one is the horizontal motion of no acceleration and the other vertical motion of constant acceleration due to gravity. The projectile motion is always in the form of a parabola which is represented as:
Browse more Topics under Motion In A Plane
- Introduction to Motion in a Plane
- Scalars and Vectors
- Resolution of Vectors and Vector Addition
- Addition and Subtraction of Vectors – Graphical Method
- Relative Velocity in Two Dimensions
- Uniform Circular Motion
y = ax + bx2
Projectile motion is calculated by a way of neglecting air resistance in order to simplify the calculations. The above diagram represents the motion of an object under the influence of gravity. It is an example of projectile motion (a special case of motion in a plane). The motion of a projectile is considered as a result:
Few Examples of Two – Dimensional Projectiles
- Throwing a ball or a cannonball
- The motion of a billiard ball on the billiard table.
- A motion of a shell fired from a gun.
- A motion of a boat in a river.
- The motion of the earth around the sun.
When a particle is projected obliquely near the earth’s surface, it moves simultaneously in the direction of horizontal and vertical. The motion of such a particle is called Projectile Motion. In the above diagram, where a particle is projected at an angle θ, with an initial velocity u. For this particular case, we will calculate the following:
- The time is taken to reach point A from O
- The horizontal distance covered (OA)
- The maximum height reached during the motion.
- The velocity at any time “ t “ during the motion.
|Components of velocity at time t||vx = v0 cosθ0
vy = v0 sinθ0–gt
|Position at time t||x = (v0 cosθ0)t
y = (v0 sinθ0)t – 1/2 gt2
|Equation of path of projectile motion||y = (tan θ0)x – gx2/2(v0cosθ0)2|
|Time of maximum height||tm = v0 sinθ0 /g|
|Time of flight||2tm = 2(v0 sinθ0/g)|
|Maximum height of projectile||hm = (v0 sinθ0)2/2g|
|Horizontal range of projectile||R = v02 sin 2θ0/g|
|Maximum horizontal range ( θ0 = 45° )||Rm = v02/g|
If any object is thrown with the velocity u, making an angle Θ from horizontal, then the horizontal component of initial velocity = u cos Θ and the vertical component of initial velocity = u sin Θ. The horizontal component of velocity (u cos Θ) remains the same during the whole journey as herein, no acceleration is acting horizontally.
The vertical component of velocity (u sin Θ) gradually decrease and at the highest point of the path becomes 0. The velocity of the body at the highest point is u cos Θ in the horizontal direction. However, the angle between the velocity and acceleration is 90 degree.
Important Points of Projectile Motion
- The linear momentum at the highest point is mu cos Θ and the kinetic energy is (1/2)m(u cos Θ)2
- After t seconds, the horizontal displacement of the projectile is x = (u cos Θ) t
- After t seconds, the vertical displacement of the projectile is y = (u sin Θ) t – (1/2) gt2
- The equation of the path of the projectile is y = x tan Θ – [g/(2(u2 cos Θ)2)]x2
- The path of a projectile is parabolic.
- At the lowest point, the kinetic energy is (1/2) mu2
- At the lowest point, the linear momentum is = mu
- Throughout the motion, the acceleration of projectile is constant and acts vertically downwards being equal to g.
- The angular momentum of projectile = mu cos Θ × h where the value of h denotes the height.
- The angle between the velocity and acceleration in the case of angular projection varies from 0 < Θ < 180 degrees.
Solved Examples for You
Question: A man can swim at a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?
Solution: Speed of the man = 4 km/h and width of the river = 1 km
Time taken to cross the river = Width of the river / Speed of the river
= 1/4 h = 1/4 × 60 = 15 minutes
Speed of the river, vr = 3 km/h
Distance covered with flow of the river = vr × t
= 3 × 1/3 = 1/4
= 3/4 × 1000 = 750 m