Question

(i) Prove that maximum horizontal range in four times the maximum height attained by projection. 

(ii) Find the angle of projection at which the horizontal range and maximum height are equal.

Answer 1

(i) Using R =\(\frac{u^2 sin 2θ}{g}\)

= \(\frac{u^2 sin^2 \theta}{2g}\)

We know, that Range is maximum when

sin 2θ = 1

i.e., θ = 45º

R_θ= 45º = \(\frac{u^2}{2g}\)

H_θ= 45º = \(\frac{u^2}{2g}.\frac{1}{2}\)= \(\frac{u^2}{4g}\)

∴\(\frac{R_(max)}{H_(max)}=4\)

R_max = 4 H_max. 

(ii) Here,

Given, Horizontal range = max. height

∴ \(\frac{u^2sin2\theta}{g}=\frac{u^2sin^2\theta}{2g}\)

or 2sinθcosθ = sin²θ/2

\(\frac{sin\theta}{cos\theta}=4\)

or θ = 75º.96