Show that maximum horizontal range of a projectie is four times the corresponding maximum height?

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Answer 1 · 2018-02-01T16:46:37

#R_max = 4H_max#

The horizontal range is maximum for #θ = 45°# and it is given by
#R_max = (u^2)/g#........................(1)

And,

the maximum height attained is # H = (u^2(sinθ)^2)/(2g)#

And here #θ = 45°#, so

#H_max = (u^2(sin45°)^2)/(2g) = u^2/(4g)# ...........................(2)

From equations (1) and (2) we get :-

#R_max = 4H_max#

NOTE :-

1) Here #θ # is the angle of projection from the Horizontal.

2) #u# is the magnitude of the velocity with which the projectile is projected.

3) #sin45°=1/(2)^(1/2)#

4) #g# is the acceleration due to Gravity.