Abaseball is hit from a height of 3.5 feet above the ground with an initial upward velocity of 48 feet per second and an initial outward velocity of 85 feet per second.
1.write the function h(t) that gives the height of the ball after time (t).
2. how long will it be before the ball hits the ground? round your answer to the nearest thousandth of a second.
3. how far will the ball travel before it hits the ground?
4. what is the maximum height the ball will reach?
5. after how long will the ball be in the air when it is on its way down at a height of 15 feet? round to the nearest thousandth of a second.

any is appreciated! i am totally stuck on this one. i know if i get a start it will but would love to know the answers to check myself too.

1. h(t) = -16t² + 48t + 3.5

2. 3.071 s

3. 261 ft

4. 39.5 ft

5. 0.334 s

Step-by-step explanation:

1. The height of a projectile h at time t is:

h(t) = -16t² + vt + s

where v is the initial vertical velocity and s is the initial height.

Here, v = 48 and s = 3.5.

h(t) = -16t² + 48t + 3.5

2. When the ball lands, h = 0.

0 = -16t² + 48t + 3.5

16t² − 48t − 3.5 = 0

t = [ -(-48) ± √((-48)² − 4(16)(-3.5)) ] / 2(16)

t = [ 48 ± √(2304 + 224) ] / 32

t = -0.071 or 3.071

The ball lands after 3.071 seconds.

3. The horizontal position of the ball is:

x(t) = 85t

x = 85 (3.071)

x = 261

The ball lands 261 feet away.

4. h(x) is a parabola, so the maximum is at the vertex.

t = -b / (2a)

t = -48 / (2×-16)

t = 1.5

h(1.5) = 39.5

The ball reaches a maximum height of 39.5 ft.

5. Find t when h = 15:

15 = -16t² + 48t + 3.5

16t² − 48t + 11.5 = 0

t = [ -(-48) ± √((-48)² − 4(16)(11.5)) ] / 2(16)

t = [ 48 ± √(2304 − 736) ] / 32

t = 0.263 or 2.737

The ball is on the way down at a height of 15 ft after 2.737 seconds.  The time left to land is:

3.071 − 2.737 = 0.334 s

your answer is 228

step-by-step explanation:

add the top and bottom bases together, then divide them by 2, then multiply by the height.

v = 1/3 a^2h

step-by-step explanation:

the volume is 1/3 a^2h.