Question

Joel, a tutor at the Math Help, decided that it would be a good
idea to drive his motorcycle off of a ramp. His height during the
jump is modeled by the equation: h=-0.2+ + 2t + 1. where h is his
height in metres, and t is time in seconds. What is the maximum
height Joel reaches on his bike, and when does he reach this height?

Answer 1

Answer:

The maximum height reached by Joel was 6 units.

Step-by-step explanation:

There seems to be a typo in the equation that models the Joel's jump. I'll assume it actually is $h(t) = -0.2*t^2 + 2*t + 1$, because that kind of movement is usually modeled with quadratic equations such as this. With this in mind, in order to find the maximum height Joel reached we need to calculate the vertex of this equation. The vertex of a quadratic equation $y(x) = a*x^2 + b*x + c$ can be calculated with the following expression:

$x_{vertex} = \frac{-b}{2*a}$

While the "y", is found by applying the x coordinate of the vertex onto the equation. In the case of our problem a = -0.2, b = 2 and c = 1. Therefore the vertex can be found at:

$t_{vertex} = \frac{-2}{2*(-0.2)} = \frac{-2}{-0.4} = 5$

Therefore the maximum height is:

$h(5) = -0.2*(5)^2 +2*5 + 1\\h(5) = -5 + 10 + 1 = 6$

The maximum height reached by Joel was 6 units.