Question

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 . 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 .

Answer 1

Answer:

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:

$s(t) = -16t^2+48t+280$

where, t is the time in sec.

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

⇒$280 = -16t^2+48t+280$

Subtract 280 from both sides we have;

⇒$0= -16t^2+48t$

⇒$0 = -16t(t-3)$

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

$0= -16t^2+48t+280$

or

$-16t^2+48t+280 = 0$

⇒$-2t+6t+35 = 0$

⇒t = -2.9441 and t = 5.9441

Since, t cannot be in negative

⇒$t = 5.9441 s$

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

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

$(3, 5.9441]$

Answer 2

First find the time that the ball is level with the top of the building on its descent. You can do this by  solving 280 = -16^2 + 48t + 280 for t. This gives t = 3 seconds . 
Then when the ball reaches the ground the  time t is obtained by solving 0 = -16t^2 + 48t + 280  This gives t = 5.94 seconds.
Answer in interval notation is  (3, 5.94].