PROJECTILE MOTION QUADRATIC WORD PROBLEMS

Problem 1 :

A rock is dropped from a 100 foot tower. The height of the rock as a function of time can be modeled by the equation: h(t) = -16t² + 100. How long does it take for the rock to reach the ground?

Solution:

h(t) = -16t² + 100

h(t) = 0

0 = -16t² + 100

16t² = 100

It takes 2.5  seconds to reach the ground.

Problem 2 :

A rock is dropped on the surface of Mars from a height of 100 feet. The height of a falling rock as a function of time since it was dropped on Mars can be modeled by the equation: h(t) = -6.5t² + 100. How long does it take for the rock to hit the surface of Mars?

Solution:

h(t) = -6.5t² + 100

h(t) = 0

0 = -6.5t² + 100

6.5t² = 100

So, it takes 3.

Problem 3 :

A ball is thrown from ground level upward at an initial velocity of 60 ft/sec. What is the ball's maximum altitude? The equation for "projectile motion" is h(t) = -16t² + 60t. 

Solution:

h(t) = -16t² + 60t

a = -16, b = 60

h(1.875) = -16(1.875)² + 60(1.875)

h(1.875) = -56.25 + 112.5

h(1.875) = 56.25 feet

So, maximum altitude is 56.25 feet.

Problem 4 :

A ball is thrown upward from the surface of Mars with an initial velocity of 60 ft/sec. What is the ball's maximum height above the surface before it starts falling back to the surface? The equation for "projectile motion" on Mars is: h(t) = -6.5t² + 60t

Solution:

h(t) = -6.5t² + 60t

h(4.6154) = -6.5(4.6154)² + 60(4.6154)

= -138.46 + 276.92

h(4.6154) = 138.46 feet

So, the maximum height is 138.46 feet.

Problem 5 :

A rock is thrown upward from ground level with an initial velocity of 50 feet/sec. When will the rock hit the ground? Projectile motion can be modeled by the equation: h(t) = -16t² + 50t.

Solution:

h(t) = -16t² + 50t

h = 0

0 =  -16t² + 50t

0 = -2t(8t - 25)

So, in 3.125 sec the rock hits the ground.

Problem 6 :

A rock thrown upward from the surface of Mars with an initial velocity of 50 feet per second. The height of a rock can be modeled by the: h(t) = -6.5t² + 50t. How long does it take the rock to fall back to the surface of Mars?

Solution:

h(t) = -6.5t² + 50t

h = 0

0 = -6.5t² + 50t

0 = t(-6.5t + 50)

So, it will take 7.612 sec to fall back to the surface of Mars.

Problem 7 :

A rock is thrown upward from the top of a 25 foot tower with an initial upward velocity of 100 ft/sec. The height of a rock above the ground as a function of time can be modeled by the equation: h(t) = -16t² + 100t + 25. How long does it take for the rock to:

a) reach its maximum height?

b) fall back to the ground?

Solution:

a) 

h(t) = -16t² + 100t + 25

h(3.125) = -16(3.125)² + 100(3.125) + 25

= -156.25 + 337.5

= 181.25 feet

Maximum height is 181.25 feet.

b) 

h(t) = -16t² + 100t + 25

0 =  -16t² + 100t + 25

a = -16, b = 100, c = 25

Time to hit the ground is 6.49 sec.

Problem 8 :

A rock is thrown downward with an initial downward velocity of 50 ft/sec from the top of a 1000 foot skyscraper. The height of a falling rock as a function of time can be modeled y the equation: h(t) = -16t² + 50t + 1000. How long does it take for the rock to hit the street below? 

Solution:

h(t) = -16t² + 50t + 1000

0 = -16t² + 50t + 1000

a = -16, b = 50, c = 1000

So, it takes 9.61 sec to hit the street below.

Problem 9 :

A rectangle with a width of (5x + 2) feet and a length of (2x - 1) feet has an area of 100 square feet. What is the rectangle's width and length?

Solution:

Given, width = 5x + 2 feet

length = 2x - 1 feet

Area of rectangle = 100 square feet

Area of rectangle = length × width

100 = (5x + 2)(2x - 1)

100 = 10x² - 5x + 4x - 2

100 = 10x² - x - 2

10x² - x - 102 = 0

By using quadratic formula,

a = 10, b = -1 and c = -102

x = 3.24

Width = 5(3.24) + 2

= 18.2 feet

Length = 2(3.24) - 1

= 5.48 feet

Problem 10 :

A rectangle with a width of (2x + 5) feet and a length of (3x - 1) feet has an area of 250 square feet. What is the rectangle's width and length?

Solution:

Given, width = 2x + 5 feet

length = 3x - 1 feet

Area of rectangle = 250 square feet

Area of rectangle = length × width

250 = (2x + 5)(3x - 1)

250 = 6x² - 2x + 15x - 5

250 = 6x² + 13x - 5

6x² + 13x - 255 = 0

By using quadratic formula,

a = 6, b = 13 and c = -255

x = 7.69

Width = 2(7.69) + 5

= 20.38 feet

Length = 3(7.69) - 1

= 22.07 feet