Question

A projectile is fired from the surface of earth with a velocity $5m{{s}^{-1}}$ of and angle ${{8}^{\circ }}$ with the horizontal. Another projectile fired from another planet with a velocity of $3m{{s}^{-1}}$ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of acceleration due to gravity on the planet is:
(A) 3.5
(B) 5.9
(C) 16.3
(D) 110.8

Answer 1

Hint: The equation of trajectory is given by$y=x\tan \theta -\dfrac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$ where $u$is velocity of projection and $\theta $ is the angle of projection. The equation of trajectory will not change when we move to the other planet because there will be vertical gravitational acceleration as well no horizontal acceleration though the value may differ. On equating the trajectory we can easily determine the value of g on another planet.

Complete step by step answer:
We know that the equation of trajectory will not change when we move to the other planet because there will be vertical gravitational acceleration as well no horizontal acceleration though the value may differ.
Also we know that equation of trajectory at earth is given by
\[y=x\tan \theta -\dfrac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\]
Equation of trajectory at another planet is given by
\[y=x\tan {{\theta }^{'}}-\dfrac{{{g}^{'}}{{x}^{2}}}{2{{u}^{'}}^{2}{{\cos }^{2}}{{\theta }^{'}}}\]
According to condition we know that
Equation of trajectory at earth = equation of trajectory at another planet
$x\tan \theta -\dfrac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }=x\tan {{\theta }^{'}}-\dfrac{{{g}^{'}}{{x}^{2}}}{2{{u}^{'}}^{2}{{\cos }^{2}}{{\theta }^{'}}}$
As given
$\theta ={{\theta }^{'}}$
$g=9.8\operatorname{m}/{{s}^{2}}$
$\operatorname{u}=5m/s\text{ }\And \text{ }u'=3m/s.$
On putting the values we get
$\dfrac{g}{{{u}^{2}}}=\dfrac{{{g}^{'}}}{{{u}^{'}}^{2}}$
On solving for ${{g}^{'}}$ we get
${{g}^{'}}=\dfrac{{{u}^{'}}^{2}\times g}{{{u}^{2}}}$
$\implies {{g}^{'}}=\dfrac{9\times 9.8}{25}$
On further solving we get
${{g}^{'}}=3.528m/{{s}^{2}}$
Hence, option (A) is correct.

Note:
A trajectory or flight path is the path that an object with mass in motion follows through space as a function of time. In classical mechanics, the mass might be a projectile or a satellite. For example, it can be an orbit, the path of a planet, asteroid, or comet as it travels around a central mass. Acceleration due to gravity depends on various factors such as height, depth as well as rotation of the planet.