Question

A projectile fired from a ship comes down 10.0 s after it
was fired and hits the water at a distance of 6000 m from
the ship. (a) What was the horizontal component of its
initial velocity? (b) At what time did it reach the highest
point of its path? (c) What was the altitude at that time?
(d) What was the vertical component of its initial
velocity

Answer 1

Ignoring air resistance...

a) To find the horizontal component of velocity, use vt = d since we know the time and horizontal distance (velocity*time = distance)
so v*10.0 = 6000, so v = 600m/s

b) It was in the air for 10 seconds, the time on the way up will equal the time on the way down (ie from t = 0 to t = 5 seconds it's on its way up, from t = 5 to t= 10 seconds it's on its way down), so the highest point will be at t = 5.00 seconds.

c) The altitude or height, h = 0.5gt^2 = 0.5*9.81*5.00^2 = 122.5m

d) vertical velocity, v = gt = 9.81*5.00 = 49.05m/s