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}} $

Show Answer

Answer:

Correct Answer: C

Solution:

[c] If t it’s the time of flight, then $ 0=vt-\frac{1}{2}( g\cos \theta)t^{2}\Rightarrow t=\frac{2v}{g}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 $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक