Question

A body is projected at a angle of thea with horizontal prove path is trochory find man height and time of flight and horizontal range

Answer 1

Answer:

Given:

An object is thrown at an angle of θ with horizontal.

To find:

1. Equation of trajectory

2. Max height

3. Time

4. Range

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.

Now, for time period:

H = u sin(θ) t - ½gt²

=> 0 = u sin(θ) t - ½gt²

=> t = {2u sin(θ)}/g

For Range:

Range = u cos(θ) × t

=> R = u cos(θ) × {2u sin(θ)}/g

=> R = u² sin(2θ)/g

For Max height:

We are putting t = {u sin(θ)}/g as at half time , we get max height.

H max.= ut - ½gt²

=> H max. = [{u sin(θ)} u sin(θ)/g] - ½gt²

=> H max. = u² sin²(θ)/2g