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| Math Central | Quandaries & Queries |
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Question from Zoraida: A penny is thrown from the top of a 26.7-meter building and hits the ground 3.39 seconds after it was thrown. The penny reached its maximum height above the ground 0.89 seconds after it was thrown.a. Define a quadratic function, h, that expresses the height of the penny above the ground (measured in meters) as a function of the number of seconds elapsed since the penny was thrown, t. b. What is the maximum height of the penny above the ground? |
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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