Astone falls freely from rest for 18 s. the acceleration due to gravity is 9.8 m/s 2. what was the stone’s displacement after that time? ans in m

-- At the end of 18sec, the stone's velocity is (9.8m/s² x 18s) = 176.4m/s downward.

-- Starting at 0 m/s and ending at 176.4m/s, its average velocity is 88.2m/s downward.

-- Falling for 18s at an average speed of 88.2m/s, it falls (88.2m/s x 18s) = 1,587.6 m .

-- Its displacement is 1,587.6 meters downward.

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