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
![a = 32 ft/s^2](/tex.php?f=a = 32 ft/s^2)
so as we know that
![a = \frac{v^2}{r}](/tex.php?f=a = \frac{v^2}{r})
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
![a_c = 32 ft/s^2](/tex.php?f=a_c = 32 ft/s^2)
![a_t = 0](/tex.php?f=a_t = 0)
now radius of curvature is given as
![r = \frac{v^2}{a_c}](/tex.php?f=r = \frac{v^2}{a_c})
![r = \frac{150^2}{32}](/tex.php?f=r = \frac{150^2}{32})
![r = 703.125ft](/tex.php?f=r = 703.125ft)
so the radius of curvature just after it is dropped is 703.125 ft