Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 11


Pls. explain equation of trajectory motion or give a suitable link.

Pls. explain equation of trajectory motion or give a suitable link.

2 years ago

Answers : (1)

25768 Points

A body projected into the space and is no longer being propelled by fuel is called a projectile.

To analyze the projectile motion we use the following concept "Resolution of two dimensional motion into two one dimension motion" as discussed earlier. Hence it is easier to analyze the motion of projectile as composed of two simultaneous rectilinear motions which are independent of each other:

(a) Along the vertical y-axis with a uniform downward acceleration 'g' and

(b) Along the horizontal x-axis with a uniform velocity forward.

Consider a particle projected with an initial velocity u at an angle θ with the horizontal x-axis as shown in figure shown below. Velocity and accelerations can be resolved into two components:

                                                A Particle Projected With an Initial Velocity u at an Angle Theta

Velocity along x-axis = ux = u cos θ

Acceleration along x-axis ax = 0

Velocity along y-axis = uy = u sin θ

Acceleration along y-axis ay = -g

Here we use different equation of motions of one dimension derived earlier to get the different parameters.

\vec{v}=\vec{v_{0}}-\vec{g}t                           …... (a) 

\vec{y}-\vec{y_{0}}=\vec{v_{0}}t-\frac{1}{2}\vec{g}t^{2}           …... (b)         

v2 = v02 – 2g (y-y0)                      …... (c) 

Total Time of Flight

When body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation b.

Therefore, 0 = (u sin θ) t - (½)gt2,

As t cannot equal to zero, then, total time of flight,

Horizontal Range

Projectile Motion

Horizontal Range (OA=X)   = Horizontal velocity × Time of flight

= u cos θ × 2 u sin θ/g

So horizontal range,

Maximum Height

At the highest point of the trajectory, vertical component of velocity is zero.

Therefore, 0 = (u sin θ)2 - 2g Hmax

So, maximum height would be,



Equation of Trajectory

Assuming the point of projection as the origin of co-ordinates and horizontal direction as the x-axis and vertical direction as the y-axis. Let P (x, y) be the position of the particle at instant after t second.

Then x = u cos θ.t and y = u sin θ.t - 1/2 gt2

Eliminating 't' form the above equations, we get,

y = x tan θ - (gx2/2u2cos2θ)

This is the equation of trajectory which is a parabola (y = ax + bx2).

2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts
Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details