Back to Course

SS2: PHYSICS - 1ST TERM

0% Complete
0/0 Steps
  1. Scalars & Vectors | Week 1
    5 Topics
    |
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    |
    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
    |
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    |
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    |
    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics
    |
    1 Quiz



  • Do you like this content?

  • Follow us

Lesson Progress
0% Complete

If an object is projected with a velocity, u, and at an angle of inclination, with the horizontal, the vertical component of the velocity is U sin θ, while the horizontal component Ux, is U cos θ

Projectile Motion 1 e1607005559711
Path of a projectile.

At the maximum height, the final vertical velocity of the projectile is 0, v = 0, then;

v = U + gt

0 = U sin θ – gt

 t = \( \frac {U\: sin \:θ}{g} \)

This is the time taken for the projectile to reach the maximum height, H, the projectile also uses the same time to return to the plane of projection.

If the time taken to reach maximum height is t1 and t2 is the time taken to return to the plane of projection, and t1 = t2 = t, then the total time of flight for the projectile to travel and return to the plane of projection is

t1 + t2

= t + t = T

T = 2t

T = \( \frac {2\:U\: sin\:θ}{g} \)

Time of flight (T) of a projectile is the time required for a projectile to return to the plane of projection.

Responses

Your email address will not be published. Required fields are marked *

error: