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PLEASE HELP!!!! I’m about to rip my hair out over this.

1. A ball is thrown into the air at an initial velocity of 18 meters per second from an initial height of 10 meters. The equation that models the path of the ball is given by h=-4.9t^2+18t+10. When does the ball reach its Maximum height?

2. A ball is thrown into the air at an initial velocity of 18 meters per second from an initial height of 10 meters. The equation that models the path of the ball is given by h=-4.9t^2+18t+10. When does the ball hit the ground?

3. A ball is thrown into the air at an initial velocity of 18 meters per second from an initial height of 10 meters. The equation that models the path of the ball is given by h=-4.9t^2+18t+10. How high is the ball at 3 seconds?

4. Logan is doing a handstand dive from a 12 meter platform. The equation that models the path of her dive is given by h=-4.9t^2+12, where t is seconds. At what height will she be after 1 second?

I absolutely hate math with all of my heart.

Respuesta :

helper27738

Step-by-step explanation:

1. The equation graph is a parabola, so the maximum height will be the vertex of the parabola. You can find the vertex coordinate t using the formula:

t = -b/2a

t = -18/2•(-4.9)

t = -18/-9.8

t = 1.84 seconds

2. The height of the ground is 0, so the balls hit the ground when the equation result is 0:

0 = -4.9t²+18t+10

Now you solve it using Bhaskara:

Δ = b² -4ac

Δ = 18² -4•(-4.9)•10

Δ = 520

t = (-b ±√Δ)/2a

t = (-18 ± √520)/2•(-4.9)

t1 = (-18 - 20.8)/-9.8

t1 = 3.96 seconds

t2 = (-18 +20.8)/-9.8

t2 = -0.28

Doesn't exist negative time, so we pick the first value found, t = 3.96 seconds

3. Now you just need to put 3 in place of t to find the result:

h = -4.9•3² +18•3 +10

h = -4.9•9 + 54 + 10

h = -44.1 + 64

h = 19.9 meters

4. You just need to put 1 in place of t to find the height:

h = -4.9•1²+12

h = -4.9+12

h = 7.1 meters

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