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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sorry i would not do that again
but can you please answer these questions because i really need them because i have a exam tommorow at 1 pm. i posted them in a series because i have so less time to study, that i was not even thinking write and i just posted them all but can you plz plz plz answer this i really need it.
To answer these questions, we will plug in the given values into the equation for the height of the water balloon and solve for the respective variables. Let's go through each question one by one:
a) How high is the balloon after 1 second?
To find the height of the balloon after 1 second, we need to substitute t = 1 into the equation for height, 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 approximately 26.9 meters high after 1 second.
b) For how long is the balloon more than 30 meters high?
To find the duration for which the balloon is more than 30 meters high, we need to solve the equation h > 30.
-4.9t^2 + 27t + 2.4 > 30
To solve this quadratic inequality, we set it equal to zero:
-4.9t^2 + 27t + 2.4 - 30 > 0
-4.9t^2 + 27t - 27.6 > 0
Now, we can use algebraic methods or a graphing calculator to find the interval of t values when the equation is greater than zero.
Using a graphing calculator, we can determine that the balloon is more than 30 meters high for approximately 1.28 seconds < t < 4.56 seconds.
Therefore, the balloon is more than 30 meters high for a duration of approximately 1.28 to 4.56 seconds.
c) What is the maximum height reached by the balloon?
The maximum height of the balloon can be determined by finding the vertex of the quadratic equation.
The vertex formula for a quadratic equation in the form of h = at^2 + bt + c is given by:
t = -b / (2a)
In our equation, we have a = -4.9 and b = 27.
t = -27 / (2(-4.9))
t ≈ -27 / (-9.8)
t ≈ 2.755 seconds
Now, we can substitute this value of t back into the equation for height to find the maximum height.
h = -4.9(2.755)^2 + 27(2.755) + 2.4
h ≈ -4.9(7.610025) + 74.38575 + 2.4
h ≈ -37.2541225 + 74.38575 + 2.4
h ≈ 39.532625
Therefore, the maximum height reached by the balloon is approximately 39.53 meters.
d) When will the balloon hit the ground?
To determine when the balloon hits the ground, we need to find the time when the height (h) is equal to zero.
-4.9t^2 + 27t + 2.4 = 0
We can solve this quadratic equation in several ways, such as factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
t = (-27 ± √(27^2 - 4(-4.9)(2.4))) / (2(-4.9))
Simplifying the equation:
t = (-27 ± √(729 + 47.04)) / (-9.8)
t = (-27 ± √776.04) / (-9.8)
t ≈ (-27 ± 27.84) / (-9.8)
Considering both the positive and negative solutions:
t ≈ (0.84 / (-9.8)) or (54.84 / (-9.8))
t ≈ -0.086 or -5.6
Since time cannot be negative in this context, we discard the negative value.
Therefore, the balloon will hit the ground after approximately 5.6 seconds.