Help with this.
A child throws a rock straight up with speed 10.9 m/s and then catches it when it falls back to its launch point. Determine the following quantities from the kinematics equations for constant acceleration. Use g = 9.8 m/s2.

What is the position of the rock after 1.001 s?

Question

Help with this.
A child throws a rock straight up with speed 10.9 m/s and then catches it when it falls back to its launch point. Determine the following quantities from the kinematics equations for constant acceleration. Use g = 9.8 m/s2. 

What is the position of the rock after 1.001 s?

matt101 · Answer

We're given initial velocity, time, and acceleration, and we want to find position. Note that at 1.001 s, the rock still hasn't reached its maximum height, when its speed is 0 m/s. We know this because if you use a=v/t to calculate t (the time it takes for the rock to reach its max height), you'll find that t=1.11 s. Even if you didn't realize that, you can use this equation:

$$d=v_0t+\frac{1}{2}at^2$$

badhi · Answer

Even for $t>1.11s$  you can jst use the matt101's equation and find the position. 

That means you dont need to be worried whether the given time (here t=1.001) is less than the time of the objects maximum height (in here t = 1.11s)