The path of a rocket is given by the function, h(t)=-15t2+150t+365, where “h” is the height in metres and “t” is the time in seconds.

Question

a) What is the maximum height of the rocket? 

b) At what time does the rocket reach its maximum height?

c) At what time does the rocket land?

d) What is the domain and range of the rocket’s function?

Answer 1

h(t)=-15t²+150t+365.

To solve using calculus, differentiate wrt t: h'(t)=-30t+150=0 when the height is maximum. So t=150/30=5 secs. (See later for solution without calculus.)

So the answer to (b) is 5 seconds.

(a) h(5)=-375+750+365=740m.

(c) The rocket lands when h(t)=0, that is, 3t²-30t-73=0 (after dividing by -5).

So, using the quadratic formula: t=(30+√(900+876))/6=12.024 seconds approx. We reject the negative result because t has to be positive.

(d) t has to be positive because negative time is for everything that happened before the rocket was launched. When the rocket falls to the ground, it ceases to be in motion so that's at t=12.024 seconds. In this problem the practical domain is [0,12.024]. h(0)=365m (starting height of the rocket) and h(5)=740m, making the practical range [0,740], because the rocket falls to the ground (h=0). (Between 365m and 740m there will be two times when the rocket is at the same height as it rises and falls.)

To solve without using calculus, we need to complete the square:

h(t)=365+150t-15t²=365-15(-10t+t²)=365-15(t²-10t+25-25),

h(t)=365-15(t-5)²+375=740-15(t-5)². The maximum value is 740 when t=5 because for all other values of t, some quantity is subtracted from 740. Therefore (a) 740m (b) 5 seconds.