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aleksley [76]
3 years ago
5

A water balloon is tossed into the air with an upward velocity of 25 ft/s. Its height h(t) in ft after t seconds is given by the

function h(t) = − 16t2 + 25t + 3. Show your work.
19. After how many seconds will the balloon hit the ground? (hint: Use the quadratic formula)

20. What will the height be at t = 1 second
Mathematics
2 answers:
sammy [17]3 years ago
7 0
19. The balloon hits the ground means -16t2+25t+3=0
After you solve the equation, there are 2 solution ~1.67 or ~-0.11 (the time can not be negative) so the solution is 1.67s
20.substitute t=1 to the function -16(1)2+25(1)+3=-16+25+3=12 ft. So the height is 12ft when t=1
Komok [63]3 years ago
6 0

Step-by-step explanation:

Given that,

A water balloon is tossed into the air with an upward velocity of 25 ft/s. Its height h(t) in ft after t seconds is given by the function as :

h(t)=-16t^2+25t+3...................(1)

(a) we need to find the time when the balloon hits the ground. When it hits the ground h(t) = 0

-16t^2+25t+3=0  

Using above equation we get the value of t, t = 1.674 seconds

So, after 1.674 seconds the balloon hits the ground.

(b) We need to find the height at t = 1 seconds. Put the value of t = 1 seconds in equation (1) as :

h(1)=-16(1)^2+25(1)+3

h = 12 feet

So, at t = 1 seconds the height of the balloon is 12 feets.

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To model and solve our situation we are going to use the equation: s= \frac{d}{t}
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1. We know that the distance between the cities is 2400 miles, so d=2400. We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, S_{a}=?. We don't know how much the westward trip takes, so t_{w}=?, and we also don't know how much the eastward trip takes, so t_{e}=?.

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
450-S_{a}= \frac{2400}{t_{w} }
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
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2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, t_{t}=11. Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so t_{t}=t_{w}+t_{e}. Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for t_{w}:
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Now we can replace t_{w} in our going westward equation to model our round trip with a system of equations:
450-S_{a}= \frac{2400}{t_{w}}
450-S_{a}= \frac{2400}{11-t_{e} } equation (1)
450+S_{a}= \frac{2400}{t_{e}} equation (2)

3. To solve our system of equations, we are going to solve for t_{e} in equations (1) (2):

From equation (1)
450-S_{a}= \frac{2400}{11-t_{e} }
11-t_{e}= \frac{2400}{450-S_{a} }
-t_{e}= \frac{2400}{450-S_{a} } -11
t_{e}=11- \frac{2400}{450-S_{a} }
t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }
t_{e}= \frac{2550-11S_{a} }{450-S_{a} } equation (3)

From equation (2):
450+S_{a}= \frac{2400}{t_{e} }
t_{e}= \frac{2400}{450+S_{a} } equation (4)

Replacing (4) in (3)
\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }
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2400(450-S_{a})=(450+S_{a})(2550-11S_{a})
1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}
11(S_{a})^{2}-67500=0
11(S_{a})^{2}=67500
(S_{a})^{2}= \frac{67500}{11}
S_{a}=+/-  \sqrt{ \frac{67500}{11} }
Since speed cannot be negative, the solution of our equation is:
S_{a}= \sqrt{ \frac{67500}{11} }
S_{a}=78.33

We can conclude that the speed of the wind is 78 mph.

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