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Please explain the how did range of a projected body is defined

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Range of a projected body is defined as the total horizontal distance it will travel in its complete motion.

Supppose an object is thrown at an angle 'A' with initial velocity 'u'.

Horizontal component of velocity: u cosA

Vertical Component: u sinA

There is no acceleration in horizontal direction, and an acceleration '-g' in vertical direction.

Therefore, Vertical height (H) = u.t + 1/2 a.t^{2 }= u.sinA t - (1/2).g.t^{2 }

Horizontal range (R) = horizontal velocity x Time (since no acceleration in horizontal direction)

= ut cosA

Now we need to find total time 't' of the motion and put it in equation of Range (R).

In complete projectile motion, the total vertical Height would be zero, as the object would hit the ground again. So time for complete motion will be: (putting H=0 in equaton (H):

0 = u.sinA t - (1/2).g.t^{2}

t = (2u sinA) / g

Putting this time 't in equation of Range (R):

R = ut cosA = (ucosA x 2u sinA) / g = ( u^{2 }2 sinA cosA ) / g = u^{2}sin2A / g

(as 2sinAcosA = sin2A)

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A projected body follows a parabolic path during its motion.

The range of a projectile is defined as the horizontal distance travelled by it from its intial position, x=y=0 to the position where it passes y=0 during its fall.

Considering, Projected velocity = u, Angle of projection = A, Acceleration due to gravity = g, Time of flight, Tf= (2u.sinA)/g ,

We get,

Horizontal Range, R=(u^{2}sin2A)/g

For Maximum R,

sin2A=1, Implies 2A=90^{o}, Implies A=45^{o}