A 1550 kg torpedo strikes a 770 kg target that is initially at rest. If the combined torpedo and target move forward with a speed of 9.44 m/s,
what is the initial velocity of the torpedo? Assume that no resistance is
provided by the water.

c. newton's third law of motion

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

the components of a vector are the parts of the vectors that are to each other.

b perpendicular

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr