Answered

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)=−4.9t2+124t+416.

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 :

Answer:

t = 28.3 seconds

1200 meters

Step-by-step explanation:

The height of the rocket after t seconds from launching is given by

h(t) = - 4.9t² + 124t +416

Now, at h = 0 the rocket will splashes down.

Hence, -4.9t² + 124t + 416 = 0

Applying Sridhar Acharya formula,

t = [tex]\frac{-124 -\sqrt{124^{2}- 4 \times(-4.9) \times416 } }{2 \times (-4.9)}[/tex] or, t = [tex]\frac{-124 +\sqrt{124^{2}- 4 \times(-4.9) \times416 } }{2 \times (-4.9)}[/tex]

Hence, t = 28.3 or, t = -2.999

Since t can not be negative, so, t = 28.3 seconds. (Answer)

Now, h = -4.9t² + 124t + 416, for h to be maximum [tex]\frac{dh}{dt}  =0[/tex]

⇒ -9.8t + 124 = 0

⇒ t =12.65 secs

Hence, [tex]h(x)_{max} =-4.9 \times(12.65)^{2}  +124 \times(12.65)+416 =1200[/tex] meters

Therefore, the rocket peaks at 1200 meters above sea level. (Answer)

Other Questions