Atoy rocket is launched vertically from 5 feet above ground level with an initial velocity of 112 feet per second. the height h after t seconds is given by the equation h(t)=-16t^2+112t+5.
a. how long will it take for the rocket to return to the ground?
b. after how many seconds will the rocket be 100 feet above the ground?

Step-by-step explanation:

a)The height h after t seconds is given by the equation h(t)=-16t^2+112t+5.

Where 5 represents 5feet above the ground and this is the height from which the rocket was launched.

The equation is a quadratic equation. The plot of this equation on a graph would give a parabola whose vertex would be equal to the maximum height travelled by the rocket.

The vertex of the parabola is calculated as follows,

Vertex = -b/2a

From the equation,

a = -16

b = 112

Vertex = - - 112/32= 3.5

So the rocket will attain maximum height at 3.5 seconds.

It will also take 3.5 seconds to reach the ground. This means it will take a total of 3.5 seconds + 3.5 seconds

= 7 seconds to hit the ground.

b) to find time it will take to be 100 feet above the ground,

-16t^2+112t-95 = 0

Using general quadratic equation formula,

a = -16

b = 112

c = -95

t = [-b+/-√b^2-4ac]/2a

t = [ -112 +/-√112^2-4×-16×-95]/16 × -2

= [-112 +/-√12544-6080]/-32

= (-112+80.4)/-32 or (-112-80.4)/-32

t = 0.9875 or t = 6.0125

So it will take 0.9875 or approximately 1 second to be 100 feet above the ground

step-by-step explanation:

this is not a question

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: