Question

derive an equation for displacement of a projectile fired at an angle theta from the ground

Answer 1

Answer:

Given:

An object is thrown at an angle θ from the horizontal.

To find:

Equation of trajectory

Concept:

We will divide the projectile motion into 2 different linear motion in x and y axes.

Calculation:

Time = T

Velocity in x axis = v cos(θ)

Velocity in y axis = v sin(θ)

Distance in x Axis =x= [v cos(θ)] × T ....(i)

Distance in y Axis

y = v sin(θ) × T - ½gT² ...........(ii)

Putting value of T in eq.(ii)

y = x × v sin(θ)/ v cos(θ) - ½g{x/vcos(θ)}²

y = x tan(θ) - gx²/2u²cos²(θ).

The derived equation is similar to a parabolic equation

y = ax - bx².

So the trajectory of a projectile is a Parabola.

Answer 2

$Given:$

An object is thrown at an angle θ from the horizontal.

To find:

Equation of trajectory

Concept:

We will divide the projectile motion into 2 different linear motion in x and y axes.

Calculation:

Time = T

Velocity in x axis = v cos(θ)

Velocity in y axis = v sin(θ)

Distance in x Axis =x= [v cos(θ)] × T ....(i)

Distance in y Axis

y = v sin(θ) × T - ½gT²_____(ii)

Putting value of T in eq.(ii)

y = x × v sin(θ)/ v cos(θ) - ½g{x/vcos(θ)}²

y = x tan(θ) - gx²/2u²cos²(θ).

The derived equation is similar to a parabolic equation

y = ax - bx².

So the trajectory of a projectile is a Parabola.