I was given the problem:
A single force acts on a 3.6 kg particle-like object in such a way that the position of the object as a function of time is given by x = 4.1t - 0.64t2 + 2.0t3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 8.1 s...
A rock is thrown vertically upward from ground level at time t = 0. At t = 1.6 s it passes the top of a tall tower, and 2.0 s later it reaches its maximum height. What is the height of the tower?
My thought process:
First find initial velocity:
v=vo+at
0=vo+(-9.8)(2)
vo=19.6 m/s...