A bullet fired from rifle attains a maximum height of 50m and crosses

1.	A bullet fired from rifle attains a maximum height of 50m and crosses a range of 200m. Find the angle of projection.
Answer» 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°.

A bullet fired from rifle attains a maximum height of 50m and crosses a range of 200m. Find the angle of projection.

Answer»

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°.