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An archer fires an arrow into the air with a velocity of 20.0m/s at an angle of 25⁰ Find:

A) The time in air before striking the ground.
B) The maximum height.
C) The horizontal distance traveled.

1 Answer

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Final answer:

The time in air before striking the ground is 2.33 seconds, the maximum height is 9.15 meters, and the horizontal distance traveled is 31.1 meters.

Step-by-step explanation:

To find the time in air before striking the ground, we can use the formula t = 2v*sin(θ)/g, where v is the initial velocity of the arrow, θ is the angle of launch, and g is the acceleration due to gravity. In this case, v = 20.0 m/s and θ = 25°. Plugging these values into the formula, we get t = 2 * 20.0 * sin(25°) / 9.8 = 2.33 seconds.

The maximum height can be found using the formula h = v^2 * sin^2(θ) / (2 * g). Plugging in the values, we get

h = 20.0^2 * sin^2(25°) / (2 * 9.8)

≈ 9.15 meters.

The horizontal distance traveled can be found using the formula d = v^2 * sin(2θ) / g. Plugging in the values, we get

d = 20.0^2 * sin(2 * 25°) / 9.8

≈ 31.1 meters.

User Esteban Brenes
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