We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance of the object from the ground. Exactly seconds after we threw the object, it’s height, (measured in feet) is ℎ= -16^2 + 256 + 1680

a.) Find ℎ when = 3.

b.) How much does the object travel during the two seconds between 5 seconds and 7 seconds?

c.) How long does it take for the object to reach a height of 2640 ft?

d.) How long does it take for the object to hit the ground?

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

(1)

Where:

- Height above the ground, measured in feet.

- Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at and :

We notice that the small object ascends in the given interval.

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that , then (1) is reduced into this second-order polynomial:

(2)

All roots of the resulting equation come from the Quadratic Formula:

and

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that , then (1) is reduced into this second-order polynomial:

(3)

All roots of the resulting equation come from the Quadratic Formula:

and

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

i can't read the graph

step-by-step explanation: