A) 5%
B) 10%
C) 15%
D) 20%
Correct Answer: A
Solution :
Let h be the maximum height attained by the projectile. Then \[h=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[\therefore \] \[\frac{R}{h}=\frac{{{u}^{2}}\sin 2\theta }{g}/\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[=\frac{2\sin \theta \cos \theta }{({{\sin }^{2}}\theta )/2}\] Therefore, \[\frac{\Delta R}{R}=\frac{\Delta h}{h}\] Hence, percentage increase in R = percentage increase in height\[h=5\text{ }%\]
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