Who has the greater acceleration rate? A runner

1 answer:

6 0

The first runner because it is very clear that accelarition depends on the time and we know that the time in this case is pretty simple

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Two stones are dropped from the edge of a 60m cliff , the second stone 1.6secon after the first . How far below the top of the c

tigry1 [53]

Answer:

The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff

Explanation:

The given parameters are;

The height of the cliff from which the stones are dropped, h = 60 m

The time at which the second stone is dropped = 1.6 seconds after the first

The distance below the top of the cliff when the distance between the two stones is 36 m = Required

We have;

The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²

For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²

For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²

t₁ = t₂ + 1.6

g = The acceleration due to gravity ≈ 9.81 m/s²

s = The distance below the cliff top

The initial velocity of the stones, u = 0

Let t represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;

s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²

s₂ = u·t + (1/2)·g·t²

u = 0

∴ s₁ - s₂ = 36 = (1/2)·g·(t + 1.6)² - (1/2)·g·t²

2 × 36/(g) = (t + 1.6)² - t² = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56

2 × 36/(9.81) = 3.2·t + 2.56

t = (2 × 36/(9.81) - 2.56)/3.2 = ≈ 1.49 s

t ≈ 1.49 s

s₂ = (1/2)·g·t²

∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9

The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.

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3 years ago

1 Which photograph of Lange's is a portrait orientation? Which photograph of Lange's is a landscape orientation? Be sure to supp

Temka [501]

Answer: Is there any photographs to choose from?

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4 years ago

The formula h = -16t squared + 64t squared gives the height, h, and time, t, of an object launched from the ground with a speed

padilas [110]

Answer:

Details in the explanation

Explanation:

Vertical Launch

When an object is thrown vertically in free air (no friction), it moves upwards at its maximum speed while the acceleration of gravity starts to brake it. At a given time and height, the object stops in mid-air and starts to fall back to the launching point until reaching it with the same speed it was launched.

We are given an expression for the height of an object in function of time t

$h = -16t^2 + 64t$