Question

A projectile is fired with a velocity v0 making angle theta with horizontal. Find an expression for time of
flight and horizontal range. Draw necessary diagram.

Answer 1

Given:

A projectile is fired with a velocity v0 making angle theta with horizontal.

To find:

Find an expression for time of flight and horizontal range.

Draw necessary diagram.

Solution:

A projectile is fired with a velocity v0 making angle theta with horizontal.

Time of flight

Resultant displacement is zero in Vertical direction.  Therefore, by using the equation of motion, we get,

s = ut - 1/2gt²

gt = 2sin θ

t = 2sin θ/g

Horizontal range

Horizontal range  = horizontal component of velocity × total flight time

R = u cos θ × 2u sin θ/g

R = u² sin 2θ/g

Answer 2

Given that,

Velocity = v₀

Angle = θ

Let us consider a object projected at an angle θ with respect to horizontal x-axis with the initial velocity u .

According to diagram,

The point A is called the point of projection, θ is the angle of projection.

The distance of AB is called horizontal range.

The total time taken by the particle from reaching point A to B is called the time of flight.

The component of velocity of object

Horizontal component of velocity $v_{x}=v_{0}\cos\theta$

Vertical component of velocity $v_{y}=v_{0}\sin\theta$

The vertical distance is

$y=v_{y}t$

$t=\dfrac{y}{v_{y}}$

$t=\dfrac{y}{v_{0}\sin\theta}$

(a). We need to find the expression of time of flight

Using second equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Where, s = vertical distance

Put the value in the equation

$y=0+\dfrac{1}{2}gt\times\dfrac{y}{v_{0}\sin\theta}$

$gt=2v_{0}\sin\theta$

$t=\dfrac{2v_{0}\sin\theta}{g}$

(b). We need to calculate the horizontal range

Using formula of distance

$D =v_{x}t$

Put the value in to the formula

$R=v\cos\theta\times\dfrac{2v_{0}\sin\theta}{g}$

$R=\dfrac{v_{0}^2\sin2\theta}{g}$

Hence, (a). The time of flight is $\dfrac{2v_{0}\sin\theta}{g}$

(b). The horizontal range is $\dfrac{v_{0}^2\sin2\theta}{g}$