Question

Prove the following: (a) For two angles of projection θ and (900 -θ) with same velocity v, (i) Range is same, (ii) Heights are in the ratio of tan2 θ:1, (iii) If the range and maximum height are equal, the angle of projection is tan-1 (4).

Answer 1

Explanation :

For two angles of projection θ and (90°-θ) with same velocity v,

• Let R₁ and R₂ be the range for angles of projection θ and (90°-θ) respectively.

Horizontal Range is given by,

 $\underline{\boxed{\bf R=\dfrac{u^2sin2\theta}{g}}}$

 $\sf R_1=\dfrac{v^2sin2\theta}{g} \\\\ \sf R_2=\dfrac{v^2sin2 (90^{\circ}-\theta)}{g} \\\\ \sf R_2=\dfrac{v^2sin(180^{\circ}-2\theta)}{g} \\\\ \sf R_2=\dfrac{v^2sin2\theta}{g} \quad [ \ sin(180^{\circ} -\alpha)=sin \alpha] \\\\ R_2=R_1$

∴ Range is same

• Let H₁ and H₂ be the heights when angles of projection are θ and (90°-θ) respectively.

The maximum height is given by,

 $\underline{\boxed{\bf H=\dfrac{u^2sin^2\theta}{2g}}}$

$\sf H_1=\dfrac{v^2sin^2\theta}{2g} \\\\ \sf H_2=\dfrac{v^2sin^2(90^{\circ}-\theta)}{2g} \\\\ \sf H_2=\dfrac{v^2cos^2\theta}{2g} \quad [ \ \bf sin^2(90^{\circ}-\alpha)=cos^2 \alpha ]$

The ratio of the heights,

$\sf \dfrac{H_1}{H_2} =\dfrac{\sf \dfrac{v^2sin^2\theta}{2g}}{\dfrac{v^2cos^2 \theta}{2g}} \\\\ \sf \dfrac{H_1}{H_2}=\dfrac{sin^2 \theta}{cos^2 \theta} \\\\ \sf \dfrac{H_1}{H_2}= \dfrac{tan^2\theta}{1} \quad \bigg[ \ \dfrac{sin^2 \alpha}{cos^2 \alpha} =tan^2 \alpha \bigg]$

• Range and maximum height are equal.

 $\sf R=H \\\\ \sf \dfrac{v^2sin2\theta}{g}=\dfrac{v^2sin^2\theta}{2g} \\\\ \sf sin2\theta=\dfrac{sin^2 \theta}{2} \\\\ \sf 2sin\theta cos\theta =\dfrac{sin \theta \times sin \theta}{2} \\\\ \sf 2cos \theta \times 2=sin \theta \\\\ 4cos \theta=sin \theta \\\\ \dfrac{sin \theta}{cos \theta}=4 \\\\ tan \theta=4 \\\\ \boxed{\bf \theta=tan^{-1} (4)}$

Hence proved!