KCET · Physics · Motion In Two Dimensions
Two objects are projected at an angle \(\theta^{\circ}\) and \(\left(90-\theta^{\circ}\right)\), to the horizontal with the same speed. The ratio of their maximum vertical heights is
- A \(\tan \theta: 1\)
- B \(1: \tan \theta\)
- C \(\tan ^2 \theta: 1\)
- D \(1: 1\)
Answer & Solution
Correct Answer
(C) \(\tan ^2 \theta: 1\)
Step-by-step Solution
Detailed explanation
We know that, maximum vertical height in projectile motion,
\(H=\frac{u^2 \sin ^2 \phi}{2 g}\)
[where, \(\phi\) is angle of projection.]
\(\Rightarrow H \propto \sin ^2 \phi \)
\(\Rightarrow \frac{H_1}{H_2}=\frac{\sin ^2 \phi_1}{\sin ^2 \phi_2}=\frac{\sin ^2 \theta}{\sin ^2(90-\theta)} \)
\(\text {[Given, } \left.\phi=\theta \text { and } \phi_2=90^{\circ}-\theta\right] \)
\(=\frac{\sin ^2 \theta}{\cos ^2 \theta}=\tan ^2 \theta \)
\(\therefore H_1: H_2=\tan ^2 \theta: 1\)
\(H=\frac{u^2 \sin ^2 \phi}{2 g}\)
[where, \(\phi\) is angle of projection.]
\(\Rightarrow H \propto \sin ^2 \phi \)
\(\Rightarrow \frac{H_1}{H_2}=\frac{\sin ^2 \phi_1}{\sin ^2 \phi_2}=\frac{\sin ^2 \theta}{\sin ^2(90-\theta)} \)
\(\text {[Given, } \left.\phi=\theta \text { and } \phi_2=90^{\circ}-\theta\right] \)
\(=\frac{\sin ^2 \theta}{\cos ^2 \theta}=\tan ^2 \theta \)
\(\therefore H_1: H_2=\tan ^2 \theta: 1\)
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