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7. State the relation between maximum range and maximum height of the projectile if the angle of projection is 45 deg
Asked by roshnibudhrani88 | 23 Mar, 2024, 05:52: PM

Let u be the initial projection velocity and θ is initial projection angle. Let g be the acceleration due to gravity .

Maximum height h reached by projectile =

h = [ u2 sin2θ ] / ( 2 g )

If angle of prjection θ = 45o , then maximum height H reached by projectile is

H = u2 / (4g)  ............................(1)

Range r of projectile is

r = ( u2 / g ) sin2θ

Maximum range R of projectile when θ = 45o is

R = ( u2 / g ) ............................( 2 )

From (1) and (2) , we get

R = 4 H

Answered by Thiyagarajan K | 24 Mar, 2024, 07:48: AM

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