. A football is punted from a height of 2.75 ft above the ground, with an initial velocity of 48 feet per
second. The motion of the football can be modeled by the equation h = -16t2 + 48t +2.75, where
h represents the height of the football, in feet, and t represents the timelin seconds.
If the football is caught 5.25 ft above the ground, approximately how long did it remain in the air?
Roundyour answer to the nearest hundredth of a second.

Step-by-step explanation:

The position function for this is given as

where h(t) is the height of the football at any time t. If the ball is caught at 5.25 feet, that h(t) value is 5.25 since we are looking for the time the ball was in the air from its initial 2.75 feet to 5.25 feet, when it was caught.

and set this equal to 0:

and factor.

When we factor we get

t = .05 and t = 2.95 in seconds.

That first time tell us how long it takes to go from the initial height of 2.75 to a point 2.5 feet higher, 5.25. To get get back down to the 5.25 foot height on the other side of the vertex, is 2.95 seconds. So the total time the ball was in the air was 3 seconds exactly.

this was already solved by me with a different person. if you look it up in the search bar at the top of the page, you will get the answer.

step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: