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guajiro [1.7K]
2 years ago
15

Two rocks, a and b, are thrown horizontally from the top of a cliff. rock a has an initial speed of 10 meters per second and roc

k b has an initial speed of 20 meters per second. how would you describe the time it takes for the rocks to reach the ground and the horizontal distance they travel? both rocks hit the ground at the same time and at the same distance from the base of the cliff. both rocks hit the ground at the same time, but rock b lands twice as far as rock a from the base of the cliff. both rocks hit the ground at the same time, but rock a lands twice as far as rock b from the base of the cliff. rock b hits the ground before rock a, and rock b lands twice as far as rock a from the base of the cliff.
Physics
1 answer:
Arte-miy333 [17]2 years ago
5 0

Both rocks hit the ground at the same time and at the same distance from the base of the cliff. Option A is correct.

<h3>What is speed ?</h3>

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while u for the final speed. its SI unit is m/sec.

If the  resistance is ignored, things falling near the Earth's surface have the same estimated acceleration due to gravity of 9.8 meters per second.

As a result, the objects' acceleration is the same, and their velocity is growing at a consistent pace.

So that both rocks hit the ground at the same time and at the same distance from the base of the cliff.

Hence, option A is correct.

To learn more about the sped, refer to the link;

brainly.com/question/7359669

#SPJ1

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Explanation:

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Two solid marbles A and B with a mass of 3.00 kg and 6.50 kg respectively have an elastic collision in one dimension. Before col
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Answer:

va = 4.79 m/s

vb = 1.29 m/s

Explanation:

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂

22.75 = 3v₁ + 6.5v₂

For an elastic collision, kinetic energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²

79.625 = 3v₁² + 6.5v₂²

Two equations, two variables.  Solve with substitution:

22.75 = 3v₁ + 6.5v₂

22.75 − 3v₁ = 6.5v₂

v₂ = (22.75 − 3v₁) / 6.5

79.625 = 3v₁² + 6.5v₂²

79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²

79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5

517.5625 = 19.5v₁² + (22.75 − 3v₁)²

517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²

0 = 28.5v₁² − 136.5v₁

0 = v₁ (28.5v₁ − 136.5)

v₁ = 0 or 4.79

We know v₁ isn't 0, so v₁ = 4.79 m/s.

Solving for v₂:

v₂ = (22.75 − 3v₁) / 6.5

v₂ = 1.29 m/s

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Answer:

I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.

Explanation:

In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:

R = V₀² Sin 2θ/g

where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.

The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:

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V₀ₓ = Horizontal Velocity

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Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.

<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>

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