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Ivenika [448]
3 years ago
5

An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t)

= -4.9t2 +19.6t + 58.8. What is the object's maximum height?
Mathematics
1 answer:
beks73 [17]3 years ago
5 0
H(t) = Ho +Vot - gt^2/2

Vo = 19.6 m/s
Ho = 58.8 m
g = 9.8 m/s^2

H(t) = 58.8 + 19.6t -9.8t^2/2 = 58.8 + 19.6t - 4.9t^2

Maximun height is at the vertex of the parabole

To find the vertex, first find the roots.

58.8 + 19.6t - 4.9t^2 = 0

Divide by 4.9

12 + 4t - t^2 = 0

Change sign and reorder

t^2 - 4t -12 = 0 

Factor

(t - 6)(t + 2) =0 ==> t = 6 and t = -2.

The vertex is in the mid point between both roots

Find H(t) for: t = [6 - 2]/2 =4/2 = 2

 Find H(t) for t = 2

H(6) = 58.8 + 19.6(2) - 4.9(2)^2 = 78.4

Answer: the maximum height is 78.4 m
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Step-by-step explanation:

<u>Given:</u>

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B(1, 3t, 2)

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Step 1: Find t.

First we have to find vectors BA and BC. We do that by subtracting the coordinates of the initial point from the coordinates of the terminal point.

In vector BA B is the initial point and A is the terminal point.

BA = OA - OB = (5-1, 3t+2-3t, 2-2) = (4, 2, 0)

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Now we can find t because we know that BA • BC = 4

BA • BC = 4

To find dot product we calculate the sum of the produts of the corresponding components.

BA • BC = (4)(0) + (2)(t) + (0)(1)

4 = (4)(0) + (2)(t) + (0)(1)

4 = 0 + 2t + 0

4 = 2t

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Now we know that:

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Step 2: Find the angle ∠ABC.

Dot product: a • b = |a| |b| cos(angle)

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|BA| |BC| cos(angle) = 4

To get magnitudes we square each compoment of the vector and sum them together. Then square root.

|BA| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{20} = 2\sqrt{5}

|BC| = \sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}

2\sqrt{5}\sqrt{5}\cos{(m\angle{ABC})} = 4

10\cos{(m\angle{ABC})} = 4

\cos(m\angle{ABC}) = \frac{4}{10}=\frac{2}{5}

m\angle{ABC} = cos^{-1}{(\frac{2}{5})}

m\angle{ABC} = 66.4218^{\circ}

Rounded to two decimal places:

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