Question

The function H(t) = -16t2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.

Part A: The projectile was launched from a height of 82 feet with an initial velocity of 60 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. (2 points)

Answer 1

@lupita1995

Answer 2

h(t) = -16^2 + 60t + 82

Answer 3

how did you get that?

Answer 4

the equation is h(t) = -16^2 + 60t + 82

Answer 5

ohhh, i posted the wrong part of the question XD
Part B: What is the maximum height that the projectile will reach? Show your work. (2 points)

Part C: Another object moves in the air along the path of g(t) = 10 + 63.8t where g(t) is the height, in feet, of the object from the ground at time t seconds.

Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points)

Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know? (2 points)

Answer 6

The position function for a projectile is given by

h(t) = - 16t² + v₀t + h₀

where h = height in ft., t = time in secs., v₀ = initial velocity in ft./sec. and h₀ = initial height in ft..

v₀ = 60 ft./sec.
h₀ = 82 ft.

A.) Position function for this problem:

h(t) = - 16t² + 60t + 82
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B.) Convert to vertex form, y = a(x - h)² + k,
where (h, k) is the vertex, h is time to max.
ht. and k is max. ht.:

y = (- 16t² + 60t) + 82
y = - 16(t² - 60/16 t) + 82
y = - 16(t² - 3,75t) + 82
y = - 16(t² - 3.75t + 3.516) + 82 - [- 16(3.516)]
y = - 16(t - 1.875)² + 82 - (- 56.25)
y = - 16(t - 1.875)² + 82 + 56.25)
y = - 16(t - 1.875)² + 138.25

Max. Ht. = 138.25 ft.
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C.) g(t) = 10 + 63.8t

If h(t) = g(t),

- 16t² + 60t + 82 = 10 + 63.8t
- 16t² + 60t - 63.8t = 10 - 82
- 16t² - 3.8t = - 72
- 16(t² + 0.2375t) = - 72
t² + 0.2375t = - 72 / - 16
t² + 0.2375t = 4.5
t² + 0.2375t + 0.0141 = 4.5 + 0.0141
(t + 0.1188)² = 4.514
t + 0.1188 = √4.514
t + 0.1188 = ± 2.125
t = - 0.1188 ± 2.125

If t = - 0.1188 + 2.125,
t ≈ 2

If t = - 0.1188 - 2.125,
t = - 0.0559

Feasible Domain: t > 0

Since t = - 0.0559 is not in the feasible domain,

t = 2

The two objects intersect after 2 secs.
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D.) The objects intersect as h(t) is decreasing
because h(t)'s vertex is at t = 1.875, which is
before the intersection.
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GRAPH: