Aball is thrown straight up from the top of a tower that is 280 ft high with an initial velocity of 48 ft/s. the height of the object can be modeled by the equation s(t) = -16t^2 + 48t + 280. determine the time(s) the ball is lower than the building. express your answer in interval notation. in two or more sentences, describe your solution method.

Interval where height of ball less than building = (3, 5.94]

Explanation;

 The equation of the ball thrown from 280 ft height is given by, , where t is the time.

This ball is thrown up so initially the displacement is 280 ft, then it increases and then it reduces after reaching maximum height. When coming down it reaches at 280 ft height after a certain time. After that time the ball is at a height than building.

So we have height = s(t) = 280 ft

So, t = 0 or t =3.

We need the 3 seconds value, after that the the ball is lower than the building.

Time to reach ground is given by substituting s(t) =0

  t = −2.9441 ( not possible)  or t = 5.9441

So at time t = 5.94 it reaches ground.

Interval where height of ball less than building = (3, 5.94]

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

idk

step-by-step explanation: