It takes a force of 80.0 n to compress the spring of a toy popgun 0.200m to "load" a 0.180-kg ball. with what speed will the ball leave the gun?

v = 9.43 m/s

Explanation:

According to Hooke's law;

F = kx ------------- eqn 1

where F = spring force = 80 N

k = spring constant = ?

x = spring stretch = 0.200 m

substitute for F and x to find k

80 = 0.2k

false br

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