Kinematics Question 647
Question: A projectile is fired with a velocity v at right angle to the slope Which is inclined at an angle $ \theta $ with the horizontal. The range of the projectile along the inclined plane is:
Options:
A) $ \frac{2{{\text{v}}^{\text{2}}}\tan \theta }{\text{g}} $
B)$ \frac{{{\text{v}}^{\text{2}}}\sec \theta }{\text{g}} $
C) $ \frac{2{{\text{v}}^{\text{2}}}\tan \theta \sec \theta }{\text{g}} $
D)$ \frac{{{\text{v}}^{\text{2}}}sin\theta }{\text{g}} $
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Answer:
Correct Answer: C
Solution:
If t is the time of flight, then $ 0=vt-\frac{1}{2}( g\cos \theta)t^{2}\Rightarrow t=\frac{2v}{g\cos \theta} . $
$ R=0+\frac{1}{2}( g\sin \theta)t^{2}=\frac{1}{2}g\sin \theta {{( \frac{2v}{g\cos \theta } )}^{2}} $
$ =\frac{2v^{2}}{g}\tan \theta \sec \theta $
BETA
