Question

derive the expression for the path time of flight and horizontal range when a body is projected from a certain height in the direction of horizontal

Answer 1

Explanation:

The Proof of all the Expression is shown below.

1.) Time of Flight ⇒

  In y -direction, .

u = usinθ

a = −g

v = 0 (at highest point)

Now, Using the first equation of motion,

v = u + at

0 = usinθ − gt

⇒ gt = usinθ

⇒t = usinθ/g

Since t is only half of the total time, therefore we can calculate total time of the journey as

T = 2t

⇒T= 2usinθ

/g

______________________________

2.) Maximum Height:

Applying Law of Conservation of Energy,

1/2mu² = mgH

⇒ 1/2 m(usinθ)²= mgH

⇒1/2 mu²sin²θ = mgH

⇒ H = u²Sin²θ/2g

_____________________

3.) Horizontal Range:

Since, the projectile is in the air for a duration T

R = u in x direction  × T

⇒R = ucosθ × 2usinθ/g

⇒R=u² 2sinθcosθ/g

∴  R=u²sin2θg

Hope it helps.