Physics (momentum)

Question

A rock is thrown vertically upward from ground level at time t =0. At t =1.5 s it passes the top of a tall tower, and 1.0 s later it reaches its maximum height.What is the height of the tower?

I was able to get the answer on my second try (when I calculated for the initial velocity at ground level when rock was thrown to final velocity of when it reaches max height; then using that answer to find out the height), however I want to understand why I was not able to get the answer (~26 meters) when I attempted to solve it in a different way:

I first made a line of A ---> B where A was ground level and B was where the rock reaches the height of the building at 1 second. I then made a second line connecting that B point to C where C is the max height of the rock thrown (Where V final is 0) at the 2.5 second mark. I then calculated B------>C to figure out the velocity at point B (with the assumption that that would work as the initial velocity for where B starts, and the final velocity for AB, where B is the same height as the building. I then calculated that Velocity and used it to find the height of the building, however I did not come to the same answer. Why was that?

Answer 1

Hi Nelson, I used both methods and got an initial velocity of 24.5 m/s (at the ground)

However, there are a couple of things in your first attempt - using the top of the building as the origin, which is actually a valid method (though I think more complicated) - the way you have described it, that might have resulted in an incorrect value.

First, the rock did not reach the top of the building at 1 second, but at 1.5 seconds. That could have been the problem. and if you failed to use the time from the top of the building to max height and instead used the total time, you would also get an incorrect value.

Here is the math for your "initial" try: v = vb + a t

setting v = 0 (top of building) and vb = the unknown, a = -9.8 m/s^2, and t = 1.0 s (given in your problem - time from top of building to max height)

0 = vb - (9.8 m/s^2)(1.0 m) --> vb = (9.8 m/s^2)(1.0 s) = 9.8 m/s

Now using vb to find v0: vb = v0 + a t

where vb = 9.8 m/s, v0 = unknown, a = -9.8 m/s^2, and t = 1.5 s (the time from ground to top of building, given in problem)

9.8 m/s = v0 - (9.8 m/s^2)(1.5 s) --> v0 = 9.9 m/s + (9.8 m/s^2)(1.5 s) = 24.5 m/s

Solving for v0 from the ground to max height: v0 = (9.8 m/s^2)(2.5 s) = 24.5 m/s

Knowing the initial velocity, you can calculate the max height of the building:

y = y0 + v0 * t + (1/2) a * t^2 where y0 = 0, y = unknown, v0 = 24.5 m/s, a = -9.8 m/s^2, and t = 2.5 s

y = (24.5 m/s)(2.5 s) - (1/2)(9.8 m/s^2)(2.5 s)^2 = 61.25 - 30.62 = 30.6 m