question_answer

A projectile is fired making an angle \[2\theta \] with horizontal with velocity\[4\,\,m/s\]. At any instant it makes an angle \[\theta \], then its velocity is:

A) \[4\cos \theta \]

B) \[4(2\cos \theta -\sec \theta )\]

C)  \[2(\sec \theta +4\cos \theta )\]

D)  \[4(\sec \theta +\cos \theta )\]

Correct Answer: B

Solution :

Key Idea: Horizontal component of velocity remains same. Let \[v\] be the velocity, when projected with angle\[\theta \], then equating the horizontal velocities in both the cases, we get                 \[v\cos \theta =u\cos 2\,\,\theta \] \[\Rightarrow \]                        \[v=\frac{u\cos 2\theta }{\cos \theta }\] where     \[\sec \theta =\frac{1}{\cos \theta }\] \[\therefore \]          \[v=u\cos 2\theta \sec \theta \] Using   \[\cos 2\theta =2{{\cos }^{2}}\theta -1,\] we get Given,          \[u=4\,\,m/s,\] we get                        \[v=4(2{{\cos }^{2}}\theta -1)\sec \theta \] \[\Rightarrow \]                      \[v=4(2\cos \theta -\sec \theta )\]