Question

A stone is thrown vertically upward with a speed of 19.0m/s .
How fast is it moving when it reaches a height of 7.37mm ?
How much time is required to reach this height?
(If there is more than one answer, separate them by a comma.)

Answer 1

Using SUVAT - http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

For 1st part of your question:
$$
u=19m/s\\
v^2=u^2+2gs\\
v=\sqrt{u^2+2gs}
$$
Where g=-9.81m/s^2 and s=0.00737m. (I converted from mm to meters)
For 2nd part of your question:
$$
v=u+gt\\
t={v-u\over g}
$$
Where v was determined in part 1.

That's it!

Answer 2

Thanks!But will this me two answers?

Answer 3

When you take the square root in computing the velocity in Part 1, you will have a positive and a negative number. On the way up, the velocity is the positive answer, on the way down, it is the negative answer.

The time, t, computed in part 1 is the time to reach the the height, s, from the ground. There will be an additional time required to reach the height again on the way down after the maximum height has been reached.

So to compute the time for it to reach the height, s, on the way down, we need to compute how long it takes the stone to get to it's maximum height, subtract out the time it took to get to s, and then double that number and then add the t we found in part II to get the total time it takes to get to the height s the second time:



The time to get to maximum height is:
$$
v=u+gt\\
t_{top}={-u\over g}
$$
The time get to s again:
$$
t+2(t_{top}-t)\\
=-t+2t_{top}
$$
Where t was computed in Part II , u is the initial velocity amd v=0, because that is the velocity at the maximum height. We multiply times 2 because it takes the same about of time to go from s to max as it does to go from max to s. We add t from part II because we need to account for the time it took to get to s the 1st time around.