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 .

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

the rocket peaks at above sea level.

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 = or, t =

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

⇒ -9.8t + 124 = 0

⇒ t =12.65 secs

Hence, meters

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

90% rotation

step-by-step explanation:

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.