A 0.15 kg ball is thrown upward and reaches a height of 7m. What will its KE be when it falls back down, just before it hits the ground?
A. 1.1 J
B. 10.3 J
C. 1.5 J
D. 15 J

please have explanation

c. magnitude and direction

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

k2s for potassium sulphide