Motion in a Plane - Result Question 29
32. A projectile is fired from the surface of the earth with a velocity of $5 ms^{-1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3 ms^{-1}$ 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 $ms^{-2}$ ) given $g=9.8 m / s^{2}$
[2014]
(a) 3.5
(b) 5.9
(c) 16.3
(d) 110.8
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Answer:
Correct Answer: 32. (a)
Solution:
- (a) Trajectory is identical for both horizontal range $=\frac{u^{2} \sin 2 \theta}{g}$
So that, $\frac{g _{\text{planet }}}{g _{\text{earth }}}=\frac{(u _{\text{planet }})^{2}}{(u _{\text{earth }})^{2}}$
Therefore, $g _{\text{planet }}=(\frac{3}{5})^{2}(9.8 m / s^{2})$
$=3.5 m / s^{2}$
