9-4 Quadratic Equations and ProjectilesA model rocket is shot at an angle into the air from the launch pad.1. The height of the rocket when it has traveled horizontally x feet from the launch pad is given by h = - .163x² + 11.43x. A. Graph this equation

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x

x= -b 2ax= -(11.43) 2(-.163)x= -11.43 -.326x 35

x 0152535455570

y 01351842001841361

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This is the x coordinate of the vertex

Sec. 9-4

B. A 75 ft. tree, 10 ft. from the launch pad, is in the path of the rocket. Will the rocket clear the top of the tree?

h = -.163(10)² + 11.43(10)

h = -16.3 + 114.3

h = 98

When the rocket is 10ft from the padit will be 98 ft. above the ground whichwill clear the 75 ft. tree.

Sec. 9-4

2. The rockets height h at t seconds after launch is given by h = -22.2t² + 133t .

a. Graph this equation.b. How high is the rocket in 2 sec.?

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-b2a

-133 2(-22.2)

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y 0 111 177 200 177 110 -1

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x = 3

From the graphits 175 ft.Using the equation its:

h= -22.2t² + 133t

-22.2(2)² + 133•2 -88.8 + 266 177.2 ft.

General Formula for the Heightof a Projectile over Time

Let h be the height (in feet) of a projectile launched with an initial velocity v feet per secondand an initial height of s feet. Then, after t seconds, h = -16t² + vt + s .

Since 16ft ≈ 4.9 meters, if the units are in metersin the formula above, then h = -4.9t² + vt + s .

1. A ball is thrown from an initial height of 10 feet with an initial velocity of 64 feet per second.a. Write an equation describing the height h in feet of the ball after t seconds.b. How high will the ball be after 3 seconds?c. What is the maximum height of the ball?

a. h = -16t² + vt + s h = -16t² + 64t + 10b. h = -16(3)² + 64(3) + 10 h = -144 + 192 + 10 h = 58 ft

c. t = - b = -(64) = 2 2a 2(-16) h = -16(2)² + 64(2) + 10 h = -64 + 128 + 10 h = 74 ft

2. An object is dropped from an initial height of 40meters.a. Write a formula describing the height of the object (in meters) after t seconds.b. After how many seconds does the object hit the ground?c. What is the maximum height of the object?

a. h = -4.9t² + vt + s h = -4.9t² + 40b. h = -4.9t² + 40 0 = -4.9t² + 40 -40 - 40 -40/-4.9 = t²√-40/-4.9 = √t² √8.2 ≈ √t² 2.9sec. ≈ t

c. t = - b = 0 = 0 2a 2(-4.9) h = -4.9(0)² + 40 h = 40m

3. Suppose a ball is thrown upward with an initial upward velocity of 30 meters per second from an initial height of 10 meters.

a Write a formula for the height in meters of the object after t seconds.b. Estimate when the ball is 40 meters high.

a. h = -4.9t² + vt + s h = -4.9t² + 30t + 10

b. 40 = -4.9t² + 30t + 10

t = - b = -30 ≈ 3 2a 2(-4.9)

1.2 sec and 4.9 sec

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