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

Question

3 years ago

10

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

the object can be modeled by the equation s(t)=−16t2+48t+280.
In two or more complete sentences explain how to determine the time(s) the ball is higher than the building in interval notation. Then, determine the time.

Mathematics

2 answers:

Pani-rosa [81] · Answer 1 · 2022-04-11T03:41:56+03:00

Pani-rosa [81]3 years ago

7 0

Answer:

i hope it helpful

thank you

Alex · Answer 2 · 2022-04-11T05:38:33+03:00

Answer:

Answer in interval notation is (3, 5.94].

Step-by-step explanation:

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].