Question

A body is projected with a velocity ‘u’ in a direction making an angle θ with the horizontal.

1. Derive the mathematical equation of the path followed.

2.  Draw the velocity-time graphs for the horizontal and vertical components of velocity of the projectile.

3. Obtain an expression for the time of flight of the projectile.

Answer 1

1. The vertical displacement of projectile at any time t, can be found using the formula.

S = ut + 1/2at²

y = usinθt – 1/2gt²

But we know horizontal displacement,

x = ucosθxt

In this equation g, θ and u are constants. Hence eq.(4) can be written in the form
y = ax + bx²

where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

2.

3. The time taken by the projectile to cover the horizontal range is called the time of flight.

Time of flight of projectile is decided by usinθ. The time of flight can be found using the formula s = ut + 1/2 at²

Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinθ, we get

0 = usinθt – 1/2gt²

1/2 gt² = usinθt

\(t = \frac{2u sin \theta}{g}\)