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 ) = − 4.9 t 2 + 37 t + 374 .

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

)

=

−

4.9

t

2

+

37

t

+

374

.

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

Harry,

To answer what time does the splashdown occur , set the equation equal to zero and solve. You will have to use the quadratic formula here.

To answer the rocket peaks ____ m above sea level use the vertex of the equation. x= -b/2a

To answer how high above sea level does the rocket get at its peak, sub the value above, the x value of the vertex, into the original equation and you will get the y value of the vertex. This will tell you how high the rocket will get at its peak.

Answer 2

about 3,8 seconds is time of maximum height

found by taking the derivative and setting it equal to zero, solving for t

-9.8t + 37 = 0

t = 37/9.8 = about 3.8

then substitute 3.8 for t into the given quadratic to find max height

set the original quadratic equation = 0 and solve for t to get time of splash down.

Had the rocket started at height zero, then splash down would have been twice the

time of max height or 7.6 seconds. But because you started higher the time of

splash down will be greater than 7.6 seconds.