Question

A stone is thrown vertically upward with a speed of 24.0m/s
(a) How fast is it moving when it reaches a height of 13.0m?
b) How much time is required to reach this height?
c) Why are there two answers to b?

Answer 1

I know there must be an equation to figure out answers a and b, but I don't see one in my notes. I know that the answer to c is because it goes up and goes back down so it passes 13.0m twice.

Answer 2

There are two ways to solve this problem, by the law of conservation of energy or using kinematics.
(A)
By conservation of energy:
KE:Kinetic energy : (1/2)mv^2
PE:potential energy : mgh
Assuming PE=0 at ground level, then total energy at ground level=mg(0)+(1/2)m(24)^2
total energy at 13m, = mg(13)+(1/2)m(v^2)
equate the two energies and solve for v.
(B) use kinemaic equation v1=v0+at
where v0=initial velocity=24 m/s
v1=velocity calculated from (A).
a=acceleration due to gravity = -9.81 m/s^2
solve for t.

Answer 3

But how do I figure out the velocity from a