__Projection: Projectile fired at an angle with Horizontal:__

** Time of flight:** It is the time taken by the projectile to return to ground, or the time for which the projectile remains in the air above the horizontal plane from the point of projection.

Time of ascent = u sinθ/g

Time of descent = u sinθ/g

And, Time of flight = Time of ascent + Time of descent

Therefore, time of flight, T = 2u sin θ/g

** Maximum Height Attend:** It is the greatest height to which a projectile rises above the point of projection. it is represented by ‘H’.

Therefore, H = u² sin² θ/2g

** Horizontal Range:** It is the distance covered by the projectile along the horizontal direction between the points of projection to the point on the ground where the projectile returns again. it is represented by ‘R’.

Therefore, R = u² sin2θ/g.

Where, u = Initial velocity in m/sec.

θ = angle of projection.

g = acceleration due to gravity = 9.81 m/sec².

__Range and Time of flight on an inclined plane:__

Consider an inclined plane which makes an angle θ0 with the horizontal direction. Suppose an object projected with some initial velocity ‘u’ at an angle ‘θ’ with the horizontal.

Let, the object strikes the inclined plane at point ‘A’ after a time ‘T’.

Suppose, x-axis is taken along the plane and y-axis is perpendicular to the plane.

The component of velocity along x-axis and y-axis are;

ux = u cos(θ – θ0) and uy = u sin(θ – θ0)

The component of ‘g’ along x-axis and y-axis are; (-g sinθ0) and (-g cosθ0)

** Time of flight:** T = {2u sin(θ – θ0)}/g. cosθ

** Range of projection on inclined plane:** R = {2u² sin(θ – θ0). cosθ}/g cos²θ0

__Notes:-__

1. For a given velocity of projection, the horizontal range will be maximum, when

sin2θ = 1, or, 2θ = 90°, or, θ = 45°

i.e. To achieve the maximum range, the object should be projected at an angle of 45°

2. Horizontal range is same for angle of projection θ and (90° – θ)

For angle of projection θ, horizontal range, R = u² sin2θ/g.

If the velocity of projection makes angle θ with the vertical, then inclination with horizontal will be (90° – θ).

If R′ be the horizontal range for angle of projection (90° – θ), then

R′ = u² sin 2(90° – θ)/g = u² sin (180 – 2θ)/g = u² sin2θ/g = R

Hence, the horizontal range is same for an angle of projection θ and (90° – θ). Thus a football kicked at 30° or at 60° will strike the ground at the same place, although when kicked at 60°, it will remain longer in air.

3. If a body is projected from a place above the surface of earth, then for the maximum range, the angle of projection should be slightly less than 45°.

4. The trajectory of a projectile is a parabola, only when the acceleration of the projectile is constant and the direction of acceleration is different from the direction of velocity of the projectile.

__Summary:__

● 1. T = total time of flight = 2u sinθ/g

● 2. Time of ascent = Time of descent = u sinθ/g

● 3. Horizontal range = R = u² sin2θ/g

● 4. Horizontal coordinate, x = (u cosθ) × t or, R/2

● 5. Vertical coordinate, y = {(u sinθ) × t} -1/2gt² or, H = u² sin²θ/2g.

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