If an object is thrown down with force from the top of a building, at a speed of 20.0 m/s, and it hits the ground with a speed of 40.0 m/s.

335 ft/s

Explanation:

In this question, we have a very important factor missing. The equation of the projectile. So, in simplifying the question I'm going to assume an equation which you can correct thereafter

s(t) = -4.9t² + v(0)t + s0

From the question, we're told that v(0) = 384 ft/s, so if we apply that, we have

s(t) = -4.9t² + 384t + s0, where s0 = 0

On differentiating the equation, we have

s(t) = v(t) = -4.9(2t) + 384(1)

v(t) = -9.8t + 384

Now, at time t = 5 seconds, the velocity v =

v(t) = -9.8(5) + 384

v(t) = -49 + 384

v(t) = 335 ft/s

Therefore, the velocity of the projectile after 5 seconds is 335 ft/s. Please leave a like if it helped you.