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Serjik [45]
2 years ago
10

A ball is thrown from the height of 35 meters with an initial downward velocity of 5m/s. The ball's height h (in meters) after t

seconds is given below:
h=35-5t-5t^2
How long after the ball is thrown it hits the ground? Round to the nearest hundredth place.
Mathematics
1 answer:
Verdich [7]2 years ago
7 0
You find this value by factoring.  Factoring finds the zeros of the function.  When the ball hits the ground, it has 0 height.  Therefore, when the function goes through the x axis, it has a y value of 0.  That means that when y = 0, the ball has hit the ground.  Use the quadratic formula to factor h(t)=-5 t^{2}-5t+35.  Notice I rearranged the terms.  When you factor that you get zeros of -3.19 and 2.19.  The two things in math that will never EVER be negative are time and distance/length.  Therefore, that tells us that the ball hits the ground 2.19 seconds after it was thrown.
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Step-by-step explanation:

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Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

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ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

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∴ 10² - x² = 17² - (21² - 42x + x²)

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Substitute at (1)

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∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

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∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

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