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. in two or more sentences explain how to determine the time the ball is lower than the building in interval notation .

As per the statement:

A ball is thrown straight up from the top of a tower that is 280 ft high with an initial velocity of 48 ft/s .

Given the height of the object can be modeled by the equation:

where, t is the time in sec.

Since, we have the height s(t) = 280 ft

⇒

Subtract 280 from both sides we have;

⇒

⇒

By zero product property we have;

t = 0 and t-3 = 0

⇒t = 0 and t = 3

so, we need 3 second,  after that the ball is lower than the building.

Now, time to reach the ground

Substitute s(t) = 0

or

⇒

⇒t = -2.9441 and t = 5.9441

Since, t cannot be in negative

⇒

so, at time t = 5.9441 sec it reaches ground.

Therefore, Interval where the height  of the ball is lower than the building is,

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

answer:   c ?

step-by-step explanation: