A projectile is fired from the surface of the earth with a velocity of 5 m / s and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 m / s 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)

Question

A projectile is fired from the surface of the earth with a velocity of 5 m / s and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 m / s 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) (A) 3.5 (B) 5.9 (C) 16.3 (D) 110.8

Tardigrade Admin · Accepted Answer

As Range =(u2 sin 2 θ/g) so g propto u2(R= same) Therefore g text planet =((3/5))2(9.8 m / s 2)=3.5 m / s 2

3.5

5.9

16.3

110.8

As Range =gu2sin2θ​ so g∝u2(R= same) Therefore gplanet ​=(53​)2(9.8m/s2)=3.5m/s2

As Range $= \frac{u ^{2} s i n 2 θ}{g}$
so $g \propto u^{2} (R =$ same)
Therefore $g_{planet} = (\frac{3}{5})^{2} (9.8 m / s^{2}) = 3.5 m / s^{2}$