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 + 118 t + 115 . 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.

A.) 24.08 seconds

B.) 825.42 metres

Step-by-step explanation:

function of time is given as

h ( t ) = − 4.9 t 2 + 118 t + 115 .

Where a = -4.9, b = 118, c = 115

Let's assume that the trajectory of the rocket is a perfect parabola.

The time t the rocket will reach its maximum height will be at the symmetry of the parabola.

t = -b/2a

Substitute b and a into the formula

t = -118/-2(4.9)

t = 118/9.8

t = 12.041 seconds

Since NASA launches the rocket at t = 0 seconds, the time it will splash down into the ocean will be 2t.

2t = 2 × 12.041 = 24.08 seconds

Therefore, the rocket splashes down after 24.08 seconds.

B.) At maximum height, time t = 12.041s

Substitute t for 12.041 in the function

h ( t ) = − 4.9 t 2 + 118 t + 115

h(t) = -4.9(12.041)^2 + 118(12.041) + 115

h(t) = -4.9(144.98) + 118(12.041) + 115

h(t) = -710.402 + 1420.82 + 115

h(t) = 825.42 metres

Therefore, the rocket get to the peak at 825.42 metres

answer: i don't understand. what are the options

step-by-step explanation:

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