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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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8 0
1 year ago
Consider a dart board made of three concentric circles of radii 1 cm, 2cm and 3 cm, respectively. The points you obtain by hitti
Vlada [557]

Answer:

The answer is 350 points

Step-by-step explanation:

The Area of a circle is πr∧2

The problem states that in the attempts, you never hit the outermost ring in the 10 attempts so we need the area of the 1cm and 2cm circles

Area of the 1st circle;  π X 1∧2 = π

Area of the Second circle; π X 2∧2 = 4π

We also need the area which is the difference between the area of the 1cm and 2cm circle

4π - π  = π

Points 50 = π/4π = 1/4

points 30 = 3π/4π = 3/4

For one attempt,

E(x) = 50 x 1/4 + 30 x 3/4

= 12.25 + 22.5

= 34.75

This is approximately 35

Therefor, 10 individual attempts will be 10 x 35 = 350

8 0
3 years ago
Read 2 more answers
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