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a projectile is fired from the surface of the earth with a velocity of `5ms^(-1)` and angle `theta` with the horizontal. Another projectile fired from another planet with a velocity of `3ms^(-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)` is given `(g=9.8 ms^(-2))`

A

16.3

B

110.8

C

3.5

D

5.9

Text Solution

Verified by Experts

The correct Answer is:
C
`y = x tan theta - (gx^(2))/(2u^(2) cos^(2) theta)`
For equal trajectories for same angle of projection
`g/u^(2) `= constant
`rArr 9.8/5^(2) = g^(.)/3^(2)`
`g. = (9.8 xx 9)/25 = 3.528 m//s^(2) = 3.5 m//s^(2)`
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