URGENT!! PLEASE HELP

In a volleyball match. Jenny serves the volleyball at 14m/s, from a height of 2.5m above the court. The height o the ball in flight can be estimated using the equation h=-4.9t^2+14t+2.5, where h is the height, in meters, and t is the time, in seconds, after she serves the ball.

If a player on the other team contacts the ball at a height of 0.5m above the court, how long does it take for the ball to reach her? Round to the nearest second.

The answer for this question is 3s, but i don't now how to do it!!
I tried replacing 2.5 (in the original relation) with 0.5m, but i got a different answer.

Please help!!! Thanks!

To find the time it takes for the ball to reach a height of 0.5m, we can set the equation h = -4.9t^2 + 14t + 2.5 equal to 0.5 and solve for t.

1. Start with the equation: -4.9t^2 + 14t + 2.5 = 0.5.

2. Subtract 0.5 from both sides to isolate the quadratic term: -4.9t^2 + 14t + 2.5 - 0.5 = 0.

3. Simplify: -4.9t^2 + 14t + 2 = 0.

4. To solve the quadratic equation, we can either factor, complete the square, or use the quadratic formula. In this case, factoring is not straightforward, so we will use the quadratic formula:

The quadratic formula is given by: t = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation (-4.9t^2 + 14t + 2 = 0).

From our equation, a = -4.9, b = 14, and c = 2.

Plug in the values into the quadratic formula and simplify:

t = (-14 ± √(14^2 - 4(-4.9)(2))) / (2(-4.9))
t = (-14 ± √(196 - (-39.2))) / (-9.8)
t = (-14 ± √(196 + 39.2)) / (-9.8)
t = (-14 ± √(235.2)) / (-9.8)

5. Calculate the value inside the square root: √(235.2) ≈ 15.33.

6. We have two possible solutions for t: (-14 + 15.33) / (-9.8) and (-14 - 15.33) / (-9.8).

a) (-14 + 15.33) / (-9.8) ≈ 1.17.
b) (-14 - 15.33) / (-9.8) ≈ -2.46.

Since time cannot be negative for this scenario, we discard the negative value.

Therefore, it takes approximately 1.17 seconds for the ball to reach a height of 0.5m.

To find the time it takes for the ball to reach a height of 0.5m above the court, we can set the equation h = 0.5 and solve for t.

The equation given is h = -4.9t^2 + 14t + 2.5. We need to substitute h with 0.5 and solve for t:

0.5 = -4.9t^2 + 14t + 2.5

To make this equation easier to solve, we can multiply through by 10 to get rid of the decimals:

5 = -49t^2 + 140t + 25

To solve this quadratic equation, we can rearrange it to the standard form (ax^2 + bx + c = 0):

-49t^2 + 140t + 20 = 0

Now, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values for a, b, and c:

t = (-140 ± √(140^2 - 4*(-49)*20)) / (2*(-49))

Simplifying further:

t = (-140 ± √(19600 + 3920)) / (-98)
t = (-140 ± √23520) / (-98)
t = (-140 ± 153.334) / (-98)

Now we can solve for both values of t:

t1 = (-140 + 153.334) / (-98) = 13.334 / (-98) ≈ -0.136

t2 = (-140 - 153.334) / (-98) = -293.334 / (-98) ≈ 2.991

Notice that one of the solutions is negative and does not make sense in the context of time. Therefore, we ignore the negative value and round the positive value to the nearest second:

t ≈ 2.991 ≈ 3 seconds

So it takes approximately 3 seconds for the ball to reach a height of 0.5m above the court.

solve for h = 0.5

-4.9t^2+14t+2.5 = 0.5

now just use the quadratic formula.