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
y
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)
=
x0123456
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
window 0 < x < 10 0 < y < 80
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