Maximum height attained by the bullet, H = 50m,
Horizontal range R = 200m,
Angle of projection 0 =?
From the equation H = \(\frac {u^2 sin^2\theta}{2g}\)
50 = \(\frac {u^2 sin^2\theta}{2g}....(1)\)
From the equation R = \(\frac{u^22sin \theta cos \theta}{g}\)
200 = \(\frac{u^22sin \theta cos \theta}{g}.....{2}\)
Divide equ (1) by equ (2), we have
\(\frac{50}{200} =\frac {sin\theta}{4cos\theta}\)
i.e \(\frac{1}{4}=\frac {1}{4}\) tanθ Thus tanθ = 1
θ = tan-1 (1.0000) = 45°.
