Question

From the top of a building 96ft. high, a ball is thrown directly upward with v=80ft/sec. Find a.) time required to reach the highest point b.) max. height attained. c.) the velocity of the ball when it reaches the ground. :)

Answer 1

x = x_0 + v_0*t + 0.5*a*t^2
v = v_0 + a*t
where v_0 is the starting velocity and v the velocity at time t and x_0 is the starting point of the object, and x is the place of the object at time t. Also, a is the acceleration which is the gravitational acceleration in this case.
Thinking about a) then... v is the velocity at the end... so might it be 0 just when the ball reaches the highest point? Also, you know the velocity at the beginning, right?
Try the rest yourself ;)

Answer 2

how did you analyzed it? any advice in solving this kind of problems?

Answer 3

@Uniquebum

Answer 4

You mean how i came up with the equations? That's fairly simple integration.
Assuming a(t) is the acceleration of the object at time t we get
a(t) = dv/dt = d^2x/dt^2. Anyhow, if you want to learn more about the theory, i suggest googling or similar.