Question

A projectile is fired from the surface of the earth with a velocity of 5ms−1 and angle θ 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 : (given = 9.8ms−2)

Answer 1

IF trajectory is same for two projectile the their maximum height are equal

                 (max height)1=(max height)2

                     u1^2sin^2θ/2g1=u2^2sin^2θ/2g2

                        u1^2/g1=u2^2/g2

                        5x5/9.8=3x3/g2

                          g2=3x3x9.8/5x5

                          g2=3.528ms-2

the value of the acceleration due to gravity on the planet is 3.528ms-2.

Answer 2

Answer:

Explanation:

FOR EARTH

U=5M/S

ANGLE=THETA

g=10m/s

FOR PLANET

U=3M/S

ANGLE=THETA

G=?

USING MAXIMUM HIEGHT FORMULA

YOU WILL GET THE ANSWER AS 3.5M/S

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