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

for any of you who need the translation here it is

step-by-step explanation:

find the angular size of the equatorial radius of the earth, if the horizontal parallax of the sun

wholesale price is the initial price before markup. retail price comes after the markup.

a price of 8 bucks that's marked up by 100%   is

formula: initial price + (percent markup in fraction/decimal)(initial price)

8+(1)(8)

8+8

16

retail price is $16.