NEET Solved Paper 2014 Question 2
Question: A projectile is fired from the surface of the earth with a velocity of 5 $ m{s^{-1}} $ and angle $ \theta $ with the horizontal. Another projectile fired from another planet with a velocity of 3 $ m{s^{-1}} $ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in $ m{s^{-2}} $ ) is (given, $ g=9.8,m{s^{-2}} $ ) [AIPMT 2014]
Options:
A) 3.5
B) 5.9
C) 16.3
D) 110.8
Show Answer
Answer:
Correct Answer: A
Solution:
If the trajectory is same for both the projectiles, their maximum height will be same.
$ {{({H _{\max }})}_1}={{({H _{\max }})}_2} $ $ \frac{u_1^{2}{{\sin }^{2}}\theta }{2g _1}=\frac{u_2^{2}{{\sin }^{2}}\theta }{2g _2} $
$ \frac{u_1^{2}}{u_2^{2}}=\frac{g _1}{g _2}\Rightarrow \frac{{{(5)}^{2}}}{{{(3)}^{2}}}=\frac{9.8}{g _2} $
$ g _2=\frac{9.8\times 9}{25}=3.5m/s^{2} $
BETA
