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Worksheet: Modeling with Parabolas

Q1:

In a baseball practice, Jacob threw the ball such that β„Ž , its height above the ground in feet 𝑑 seconds after he released it, could be modeled by the quadratic equation

Write an equation that could be solved to find the times at which the ball was at a height of 33 feet above the ground.

  • A βˆ’ 1 6 𝑑 + 4 6 𝑑 + 5 = 0 2
  • B βˆ’ 1 6 𝑑 + 4 6 𝑑 + 5 = βˆ’ 3 3 2
  • C βˆ’ 1 6 𝑑 βˆ’ 4 6 𝑑 + 5 = 0 2
  • D βˆ’ 1 6 𝑑 + 4 6 𝑑 + 5 = 3 3 2

For how long was the ball 33 feet above the ground?

  • A 1 1 8 s
  • B 2 4 5 s
  • C 3 8 5 s
  • D 2 1 8 s
  • E 1 1 5 s

Q2:

A missile is launched vertically upward at a speed of 362.6 m/s.

Its height above the point from which it was launched can be found using where 𝑠 m is the height above the launch point, 𝑒 m/s is its speed when it was launched, and 𝑑 s is the time after it was launched. Find the times at which the missile is at a height of 6 629.7 m above the launch point.

  • AThe missile will reach that height twice: after 𝑑 = 3 4 seconds and again after 𝑑 = 4 2 seconds.
  • BThe missile will reach that height twice: after 𝑑 = 3 3 seconds and again after 𝑑 = 4 2 seconds.
  • CThe missile will reach that height twice: after 𝑑 = 3 4 seconds and again after 𝑑 = 4 1 seconds.
  • DThe missile will reach that height twice: after 𝑑 = 3 3 seconds and again after 𝑑 = 4 1 seconds.

Q3:

A shell was fired from a mortar along a trajectory described by the equation 𝑦 = 0 . 1 9 + 0 . 3 1 π‘₯ βˆ’ 0 . 5 π‘₯ 2 , where 𝑦 is the height of the shell above the ground in kilometres when it has travelled a horizontal distance of π‘₯ kilometres. Find the horizontal distance covered by the shell before it hit the ground.

Q4:

The height of a missile above its launch point can be found using where 𝑠 m is the height above the launch point, 𝑒 m/s is the vertical launch speed, and 𝑑 s is the time after launch. A missile is launched vertically upwards with a speed of 49 m/s. At what times will it be 44.1 m above its launch point?

  • A2.5, 11.5
  • B1, 10
  • C1, 8
  • D1, 9
  • E0.5, 8.5

Q5:

The height above the ground of a golf ball can be found using the equation 𝑦 = βˆ’ 1 6 . 1 𝑑 + 1 1 3 𝑑 + 9 2 , where 𝑦 is the height in feet and 𝑑 is the time in seconds after the ball was struck. How long will it take for the ball to reach the ground? Give your answer to 2 decimal places.

Q6:

A diver jumped from a board 15 meters above the surface of the water. The height of the dive can be found using the equation , where after seconds, the diver is meters above the water. Determine the time it took for the diver to reach the water, giving your answer to 2 decimal places.

  • A 2.67 s
  • B 2.29 s
  • C 4.59 s
  • D 1.33 s

Q7:

After 𝑑 seconds, the height, 𝑆 , of a diver above the surface of the pool is given by the equation 𝑆 = βˆ’ 4 . 9 𝑑 βˆ’ 0 . 7 𝑑 + 9  . How long does it take for the diver to reach the water?

  • A 1 3 7 s
  • B 7 9 s
  • C 7 1 0 s
  • D 1 2 7 s
  • E 9 1 0 s

Q8:

A sprinkler is used to water a garden. The height β„Ž , in meters, of the jet of water a horizontal distance π‘₯ meters from the sprinkler can be found using the quadratic equation

Write an equation that will give the distance, in meters, between the sprinkler and the point where the jet hits the ground.

  • A π‘₯ βˆ’ π‘₯ = 0 2
  • B π‘₯ + π‘₯ 4 = 0 2
  • C π‘₯ + π‘₯ = 0 2
  • D π‘₯ βˆ’ π‘₯ 4 = 0 2
  • E π‘₯ βˆ’ π‘₯ 4 = 0 2

How far from the sprinkler does the jet hit the ground?

Q9:

The arch of a curved iron gate with side length 5 m can be represented by the function 𝑓 ( π‘₯ ) = π‘Ž ( π‘₯ βˆ’ 4 ) + 9 2 . Find the value of π‘Ž .

  • A 1 4
  • B 7 8
  • C βˆ’ 7 8
  • D βˆ’ 1 4

Q10:

The arch of a curved iron gate with side length 3 m can be represented by the function 𝑓 ( π‘₯ ) = βˆ’ 3 4 ( π‘₯ βˆ’ 2 ) + 6 2 . Find the maximum height and width of the gate.

  • AThe maximum height is 4 meters and the width is 6 meters.
  • BThe maximum height is 6 meters and the width is 2 meters.
  • CThe maximum height is 2 meters and the width is 6 meters.
  • DThe maximum height is 6 meters and the width is 4 meters.

Q11:

A projectile is launched vertically upward at a speed of 120 ft/s from a 5-foot-high pad. Its height β„Ž , 𝑑 seconds after it is launched, is modeled by the formula β„Ž = βˆ’ 1 6 𝑑 + 1 2 0 𝑑 + 5 2 . Find, to the nearest hundredth of a second, the times at which the projectile is at a height of 100 ft.

  • A 0.75 s, 8.26 s
  • B 0.95 s, 6.55 s
  • C 6.60 s
  • D 0.90 s, 6.60 s
  • E 0.90 s, 0.95 s

Q12:

A toy rocket’s height, 𝑦 feet above the ground, is described by the equation 𝑦 = βˆ’ π‘₯ + 1 1 π‘₯ βˆ’ 3 0 2 , where π‘₯ feet is the horizontal distance traveled. What is the distance between the point it takes off and the point it lands?

Q13:

At an athletics competition, the flight path of the discus thrown by one of the competitors can be described by the equation 𝑦 = βˆ’ 0 . 0 9 5 π‘₯ + 2 . 1 π‘₯ + 2 . 1 2 , where 𝑦 is the height of the discus above the ground in meters when it has traveled a horizontal distance of π‘₯ meters. How far does the discus travel horizontally before it hits the ground for the first time?

  • A 46.01 m
  • B 0.96 m
  • C 1.92 m
  • D 23.06 m
  • E 22.94 m

Q14:

A garden is watered using a hose. The path of the water can be described by the equation 𝑦 = βˆ’ 0 . 3 1 π‘₯ + 1 . 9 π‘₯ + 1 2 , where 𝑦 is the height of the water above the ground in meters when it is a horizontal distance of π‘₯ meters from the hose. What is the maximum value of π‘₯ to the nearest cm?

  • A 1 323 cm
  • B 49 cm
  • C 555 cm
  • D 662 cm

Q15:

A child threw a ball vertically upwards with a speed 𝑒 m/s. The ball’s height above the ground 𝑑 seconds after it was thrown was 8 meters. The values of 𝑒 and 𝑑 satisfy the equation βˆ’ 5 𝑑 + 𝑒 𝑑 + 5 = 8 2 .

If the ball reached 8 m after a quarter of a second, at what speed was it thrown?

If the ball’s initial speed was 16 m/s, for how long was it higher than 8 m?

  • A 2 4 5 s
  • B 2 4 5 s
  • C 2 3 5 s
  • D 3 1 8 s
  • E 1 3 5 s