NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t) = − 4.9t^2 + 334t + 255.

Question

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t) = − 4.9t^2 + 334t + 255.

Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

The rocket splashes down after __________ seconds.

How high above sea-level does the rocket get at its peak?

The rocket peaks at ______________ meters above sea-level.

Answer 1

(a) The rock splashes down when h(t) = 0. Set h(t) = 0 and solve for t using quadratic formula.

0 = − 4.9t^2 + 334t + 255

t = -0.75, 68.92

(Throw away the negative answer, since time can't be negative.)

t = 68.92 sec

(b) The rock is at it's peak when the slope of it's trajectory is zero. This trajectory is also the tangent line, so finding when the slope of the tangent line is zero is the goal. To do this, take the derivative of h(t) and set h'(t) = 0.

h'(t) = -9.8t + 334

0 = -9.8t + 334

t = 34.08 sec

Use this time to find the height of the ball h(t), so h(34.08 sec).

h(t) = − 4.9(34.08)^2 + 334(34.08) + 255

h(t) = 5946.6m

Answer 2

Part 1 We are given that the height of the rocket above sea-level as a function of time is equal to

h(t) = -4.9 t^2 + 334 t + 255

The time the rocket splashes down will be the largest positive root to the function h(t), namely the largest time t_s such that h(t_s) = 0. Thus we need to solve for t_s in the equation

-4.9 (t_s)^2 + 334 t_s + 255 = 0

This can be solved using, say, the quadratic formula. This gives us that t_s can take on the values -0.755 seconds or 68.918 seconds. We want the largest of the two roots, which implies that t_s = 68.918 seconds.

Part 2 To tackle the latter question, we need to find when a maximum in the height takes place. We can do this readily via calculus by finding where the slope of h(t) is equal to 0. Taking the derivative, we find that the slope at any time t is equal to

h'(t) = -9.8 t + 334

We can find the time the rocket hits its peak altitude, t_p, by solving h'(t_p) = 0. Thus, we have that

0 = -9.8 t_p + 334

which implies that t_p = 34.08 seconds is the time the rocket reaches its peak height above sea level, completing the problem.