rocket maximum question

Question

The path of a rocket is given by the function, h(t) = -3t² + 30t + 73, where 'h' is the height in meters and 't' is the time in seconds

a. what is maximum height of rocket

b. at what time does rocket reach maximum height?

Answer 1

The function, h(t), has a "2" as its highest exponent. This makes its graph a parabola. Since the leading coefficient (the number in front of the t²) is less than 0, this is an upside down U-shaped graph with the vertex (highest or lowest point) as the maximum.

Step 1: Find the x-coordinate of the vertex (Time = t would be the substitute for the x-coordinate)

t = -b/(2a) = 30/[2·(¯3)] = -30/(-6) = 5

Time to reach max height is 5 seconds

Step 2: Plug in time to h(t) to find max height

h(5) = -3·(5)² + 30·(5) + 73

= -3 ·25 + 150 + 73

= -75 + 150 + 73

= 148 meters

The maximum height is 148 meters

Answer 2

Note this could also be solved by differential calculus.

h'(t) = -6t + 30.

For maxima and minima, set h'(t) = 0, so t = 5. It is a maximum, since h"(t) = -6, and maxima occur when the 2nd derivative is < 0.

Note also that the plug-in can be done mentally, without calculators, realizing the first term of -75 can be combined with 73 to give -2, and subtracting from 150 gives 148.

Answer 3

Method 1:

Knowledge you show know: (a+b)²= a²+2ab+b², and (a-b)²= a^2-2ab+b²

h(t)=-3t²+30t+73

= -3(t² -10t -73/3)

= -3(t²-2*5*t+5² -73/3 - 5²)

= -3 ((t-5)² -148/3)

=148 - 3(t-5)²

^{a. Maximum height: 148}

^{b When the rocket reaches maximum hight: t-5=0, so t=5}

Method 2:

Remember the formula to find the maximum or minimum point of quadratic function;

h(t)=-3t²+30t+73

a=-3

b=30

c=73

When the rocket reaches maximum, t=(-b)/(2a)= 5

Substitute 5 to the equation h(t)=-3t²+30t+73, we have h=148.

Answer 4

The maximum/minimum of a quadratic equation is on the axis of symmetry. First, we can find that t-value using the axis of symmetry formula, and then plug that back in to get height of that vertex.

1. axis of symmetry formula: t= -b/2a. For this problem, a= -3, b= 30, and c= 73. So the axis of symmetry is -30/(2(-3)) = -30/-6 = 5. So the maximum occurs at t=5 seconds. That is actually also the answer to question b.
2. plug in that t-value to find the maximum height, h(t). h(5) = -3(5)² + 30(5) + 73. So h(5)= 148 meters. The maximum height is 148 meters.