A water balloon is shot into the air so that its height h, in metres, after t seconds is

h = —4.9t^2 + 27t + 2.4

a)How high is the balloon after 1 s?
b)For how long is the balloon more than 30 m high?
c)What is the maximum height reached by the balloon?
d)When will the balloon hit the ground?

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thnx a lot now i fully understand

a) To find the height of the balloon after 1 second, we simply substitute t = 1 into the equation:

h = -4.9(1)^2 + 27(1) + 2.4
h = -4.9 + 27 + 2.4
h ≈ 24.5 meters

Therefore, the balloon is approximately 24.5 meters high after 1 second.

b) To find the duration during which the balloon is more than 30 meters high, we set the equation equal to 30 and solve for t:

-4.9t^2 + 27t + 2.4 = 30

Rearranging the equation:

-4.9t^2 + 27t + 2.4 - 30 = 0
-4.9t^2 + 27t - 27.6 = 0

Using the quadratic formula:

t = (-27 ± √(27^2 - 4(-4.9)(-27.6)))/(2(-4.9))

t ≈ 5.302 seconds and t ≈ -1.205 seconds

Since time cannot be negative, we can discard t ≈ -1.205 seconds. Therefore, the balloon is more than 30 meters high for approximately 5.302 seconds.

c) The maximum height of the balloon can be found by recognizing that it occurs at the vertex of the parabolic equation. The x-coordinate of the vertex can be found using the formula:

t = -b / (2a)

In this case, a = -4.9 and b = 27, so:

t = -27 / (2(-4.9))
t ≈ 2.755 seconds

Substituting t = 2.755 into the equation, we can find the maximum height:

h = -4.9(2.755)^2 + 27(2.755) + 2.4
h ≈ 38.865 meters

Therefore, the maximum height reached by the balloon is approximately 38.865 meters.

d) When the balloon hits the ground, its height h will be equal to 0. We can set the equation to 0 and solve for t:

-4.9t^2 + 27t + 2.4 = 0

Using the quadratic formula:

t = (-27 ± √(27^2 - 4(-4.9)(2.4)))/(2(-4.9))

Simplifying:

t ≈ 5.381 seconds and t ≈ 0.348 seconds

Since the balloon is launched upwards, we can discard the negative value. Therefore, the balloon will hit the ground after approximately 5.381 seconds.

Please note that all these calculations are approximate due to rounding.

Of course! I'll help you solve each part of the problem step by step. Here's how you can find the answers:

a) To find the height of the balloon after 1 second, we can substitute t = 1 into the given equation for h:
h = -4.9(1)^2 + 27(1) + 2.4
h = -4.9 + 27 + 2.4
h = 24.5 + 2.4
h = 26.9 meters

Therefore, the balloon is 26.9 meters high after 1 second.

b) To find how long the balloon is more than 30 meters high, we need to set the height equation greater than 30 and solve for t:
-4.9t^2 + 27t + 2.4 > 30

We can rewrite the equation as:
-4.9t^2 + 27t - 27.6 > 0

Using polynomial factoring or the quadratic formula, we can solve the equation to find the values of t. This gives us two values: t ≈ 0.891 and t ≈ 6.066.

Therefore, the balloon is more than 30 meters high for approximately 0.891 seconds and approximately 6.066 seconds.

c) The maximum height reached by the balloon corresponds to the vertex of the parabolic equation. The x-coordinate of the vertex is given by the formula:
t = -b / (2a)

In this case, a = -4.9 and b = 27. Plugging these values into the formula:
t = -27 / (2*(-4.9))
t ≈ 2.755

Now, substituting this value back into the original equation for h:
h = -4.9(2.755)^2 + 27(2.755) + 2.4
h ≈ 37.565

Therefore, the maximum height reached by the balloon is approximately 37.565 meters.

d) To find when the balloon hits the ground, we set the height equation equal to zero:
-4.9t^2 + 27t + 2.4 = 0

Again, using factoring or the quadratic formula, we can solve for t. This gives us two solutions: t ≈ -0.203 and t ≈ 5.553.

Since time cannot be negative in this context, the balloon hits the ground approximately 5.553 seconds after being shot into the air.

I hope this helps you find the answers you need! Let me know if there's anything else I can assist you with.

a) just sub t=1 into the function

b) set -4.9t^2 + 27t + 2.4 = 30
and solve. You should get two solution for t
The balloon will be above 30 m for all values of t between those two solutions

c) find the vertex, the max height is the h coordinate of the vertex
Use this method:
for y = ax^2 + bx + c, the x of the vertex is -b/(2a)
once you have that value, sub it back in to get the y.
(of course you are dealing with t instead of x, and h instead of y)

d) set —4.9t^2 + 27t + 2.4 = 0 and solve for t
One of the values of t will be negative, that would be an extraneous root.