Heidi T. answered • 6d

MA in Mathematics, PhD in Physics with 7+ years teaching experience

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