Solving Simple Kinematics: Person Throwing Rock from 40m High

[tony873004]

A person stands on the roof of a building, 40 m high. He throws a rock directly up at 10 m/s. How long until it hits the ground?

I know how to do this in 2 steps: compute height that the rock rises to (~5 meters), then compute time for an object to free fall from 0 m/s to the ground ~45 meters below.

But isn't there a way to do it in one step?

 

[HallsofIvy]

You certainly should be able to get the formula for height as a function of time: you probably use it or some version of it to get the height the rock rises to.
It is h= -4.9t²+ 10t+ 40.

Now set h= 0 and solve for t (you will get two solutions- only one is positive).

 

[wais]

Yes, there is a way to solve this problem in one step by using the equation of motion: height = initial height + initial velocity * time + 1/2 * acceleration * time^2. We can rearrange this equation to solve for time, which gives us the formula: time = (final height - initial height) / (initial velocity * acceleration). Plugging in the given values, we get: time = (0 - 40) / (-10 * (-9.8)) = 4.08 seconds. Therefore, it will take approximately 4.08 seconds for the rock to hit the ground. This method allows us to solve the problem in one step by considering the entire motion of the rock from its initial position to its final position.