



 
Hi, The expression I remember for motion along a line is \[s = u t + \frac12 a t^2\] where $s$ is the displacement, $u$ is the initial velocity, $t$ t is time and $a$ is acceleration due to gravity. I am going to measure positive as upwards in meters, $s = 0$ at the top of the building and time in seconds. Thus the penny will hit the ground when $s = 26.7$ meters. and $t = 3.39$ seconds. I also know that $a = 9.80$ meters per second squared.This allows you to solve for $u.$ Since the penny takes $0,89$ seconds to reach its maximum height $u$ must be positive. I also know that $s = 0$ twice, once at $t = 0$ and again on its way down when it passed the top of the building. Since \[0 = ut + \frac12 a t^2= t( u + a t)\] At what time does the penny pass the top of the building on its way down? At what time does the penny reach its maximum height? Penny  


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