A toy rocket is launched from the top of a building 133 feet tall at an initial velocity of 203 feet per second. a) Give the function that describes the height of the rocket in terms of time t.

Question

A toy rocket is launched from the top of a building 133 feet tall at an initial velocity of 203 feet per second. a) Give the function that describes the height of the rocket in terms of time t. b) Determine the time at which the rocket reaches its maximum height, and the maximum height in feet. c) For what time interval will the rocket be more than 738 feet above ground level? d) After how many seconds will it hit the ground?

Answer 1

Object that experience only the force of gravity can be modeled using a quadratic function. Their position, with respect to time will fit this model:

Let h(t) be the height (h) measured in feet as a function of time (t) measured in seconds then

h(t) = (initial height) + (initial velocity)•t + 1/2(-32)•t² or, in this case:

h(t) = 133 + 203t - 16t²

This is a parabola, so the time at maximum height is the vertex of the parabola. To find the value of "t" at that vertex, use t = -b/(2a) where h(t) = at² + bt + c meaning a = -16, b = 203, and c = 133

So t (at the vertex) = -203/[2(-16)] = -203/-32 = 6.34 seconds

To find the time interval above 738 feet, you can write

738 < 133 + 203t - 16t²

To solve this, PRETEND the "<" is an "=" and solve to find the boundary points:

738 = 133 + 203t - 16t²

-16t² + 203t - 605 = 0

Use the quadratic equation:

t = [-b ± √(b² - 4ac)]/(2a) where a = -16, b = 203, and c = -605

t = [-203 ± √(203² - 4(-16)(-605))]/(2(-16))

t = [-203 ± √2489]/(-32)

t = -153.11/-32 or t = -252.89/-32

t = 4.79 seconds or t = 7.90 seconds

These are the boundaries where the height is either above 738 feet or below 738 feet. A little thinking will help you determine that between t = 4.78 and t = 7.90 is when the toy is above 738 feet. If you want to be sure, you can try a value in between 4.78 and 7.90 such as t = 6 and plug that time in to see what the height is. If you do, you find that h(6) = 775 feet which verifies that this is the correct interval. You could write the interval as an inequality such as "the height is above 738 feet for the time interval 4.78 < t < 7.90

To find the time when the rocket will hit the ground, you must realize that the ground is height zero so:

0 = 133 + 203t - 16t²

Use the quadratic formula again to solve for "t"

t = [-203 ± √(203² - 4(-16)(133))]/(2(-16))

I'll let you do the rest.