After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm . the explorer finds that the pendulum completes 101 full swing cycles in a time of 140 s . what is the value of the acceleration of gravity on this planet?
express your answer in meters per second per second.

g = 10.0 m/s/s

Explanation:

For a simple pendulum, provided that the angle between the lowest and highest point of  his trajectory be small, the oscillation period is given by the following expression:

T = 2π √L/g , where L = pendulum length, g= accelleration of gravity.

We can also define the period, as the time needed to complete a full swing, so from the measured  values, we can conclude the following :

T = 140 sec/ 101 cycles = 1.39 sec

Equating both definitions for T, we can solve for g, as follows:

g = 4 π² L / T² = 4π². 0.49 m / (1.39)² = 10.0 m/s/s

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft