A penny is thrown from the top of a 28.9-meter building and hits the ground 2.71 seconds after it was thrown. The penny reached its maximum height above the gro

Question

A penny is thrown from the top of a 28.9-meter building and hits the ground 2.71 seconds after it was thrown. The penny reached its maximum height above the ground 0.56 seconds after it was thrown.

Define a quadratic function, 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, h(t)

Answer 1

h(t) = (-1/2)gt² + v₀t + h₀

 

g = 9.8 m/s²

h₀ = 28.9 m

v₀ = ?

 

h(t) = -4.9t² + v₀t + 28.9  [h in meters, t in seconds]

 

Since it hits the ground at 2.71 s,

 

h(2.71) = 0

 

0 = -4.9(2.71)² + v₀(2.71) + 28.9

 

Solve for v₀ and get

 

v₀ ≅ 2.61 m/s

 

h(t) = -4.9t² + 2.61t + 28.9

 

I suspect there may be a typo in one or more numbers in the problem statement.  The equation above is correct based on launch height and time to hit ground.  However, the time to maximum height should be the time at the vertex of the height vs. time parabola, which is at

 

t = -2.61/[2(-4.9)] s ≅ 0.266 s ≠ 0.56 s

 

Would you double-check the numbers in the original problem statement?