Help An object is thrown straight down from the top of a 300ft building with an initial velocity of -20 feet per second What is its velocity after 2 seconds and what it velocity after falling 150ft?

Answer 1

Given that dv/dt = a, you can take the integral of a, and get...

vt = a t + C. Substitute v0 for C, and you get vt = a t + v0

Assuming that the acceleration due to gravity is about -32 fps2, then the velocity after 2 seconds of an object down with an initial velocity of -20 fps is -80 fps.

To solve the second part of the question, "what is the velocity after falling 150 f?", you need to integrate again, and get...

xt = 1/2 a t2 + v0 t + C. Substitute x0 for C, and you get xt = 1/2 a t2 + v0 t + x0

Solve for t, with xt = 150, (300-150), v0 = -20, and x0 = 300, using the quadratic equation, and you get t = 2.28 s. (You also get t = -1.03 s, but you can ignore that because it is the solution for time prior to the release point.)

Go back to the first integral and plug in t = 2.28 s, and you get v = 92.9 fps.

This answer assumes no resistance from velocity in air. At 92.9 fps, you might get some effect, depending on the shape and density of the object, so the real answer is less than stated.