h=2(t-5)^2-50

Question

The distance a person falls on a bungee chord in metres, h may be modelled with time, t in seconds according to: h=2(t-5)^2-50 What is the greatest distance a person will reach on the bungee chord from their starting position?

Answer 1

Hi Rachel,

 

It will be at t=5 sec. The distance at t=5 will be 50 meters.

At t=1, h=-18

t=2, h=-32

t=3, h=-42

t=4, h=-48

t=5, h=-50

t=6, h=-48

 

You can find this by finding the vertex x coordinate of the qquadratic equation.

h=2(t-5)^2-50

= 2(t^2-10t+25)-50

= 2t^2 - 20t +50 - 50

= 2t^2 - 20t 

 

Vertex x coordinate is equal to -b/(2a)

a = 2, b = -20

= -(-20)/(2*2)

= 5

 

You can also solve this using derivatives. It will be where the function has slope zero but I don't think this is covered in Algebra.

Answer 2

The equation is in vertex form. The vertex is (5, -50). That is the maximum. Not sure how the maximum height could be a negative. Check your data.