A girl shoots an arrow from the top of a cliff. The arrow is initially at a point 20 meters above the level field below. The arrow is shot at an angle of 30 degrees above the horizontal with a speed of 39.2m/s. How far out from the base of the cliff will the arrow land?
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To find the horizontal distance traveled by the arrow, we need to calculate the time the arrow spends in the air. We can use the kinematic equation:
y = y0 + v0y * t - 1/2 * g * t^2,
where y is the vertical position, y0 is the initial vertical position, v0y is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
In this case:
y0 = 20 m,
v0y = v0 * sinθ, where θ is the launch angle,
v0 = 39.2 m/s,
θ = 30 degrees.
First, calculate v0y:
v0y = 39.2 m/s * sin(30°).
Since we need to find the time it takes for the arrow to reach the ground, we can set y = 0.
0 = 20 m + (39.2 m/s * sin(30°)) * t - 1/2 * 9.8 m/s^2 * t^2.
Rearranging the equation, we get a quadratic equation:
-4.9 t^2 + 19.6 * t + 20 = 0.
Solving this quadratic equation will give us the time the arrow spends in the air.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = -4.9, b = 19.6, and c = 20, we can calculate the values for t.
We have two solutions for t: t1 and t2. Since we are interested in the positive time, we can disregard the negative value.
Now, to find the horizontal distance traveled by the arrow, we can use the equation:
x = v0x * t,
where v0x is the initial horizontal velocity.
v0x = v0 * cosθ, where θ is the launch angle.
θ = 30 degrees,
v0 = 39.2 m/s.
Calculate v0x: v0x = 39.2 m/s * cos(30°).
Finally, substitute the values for v0x and t1 or t2 into the equation x = v0x * t to find the horizontal distance traveled by the arrow.
x = v0x * t, where t is the positive solution from the quadratic equation.
Therefore, the arrow will land at a horizontal distance x meters from the base of the cliff.
To find out how far out from the base of the cliff the arrow will land, we can use the principles of projectile motion. The horizontal distance traveled by the arrow can be determined by calculating the horizontal component of its velocity and the time it spends in the air.
First, we need to find the initial horizontal velocity (Vx) and the initial vertical velocity (Vy) of the arrow.
Vx = velocity * cos(angle)
Vy = velocity * sin(angle)
Given that the velocity is 39.2 m/s and the angle is 30 degrees, we can calculate the horizontal and vertical components:
Vx = 39.2 * cos(30)
Vy = 39.2 * sin(30)
Vx = 39.2 * 0.866
Vy = 39.2 * 0.5
Vx ≈ 33.955
Vy ≈ 19.6
Next, we need to calculate the time it takes for the arrow to reach the ground. We use the vertical component of its velocity for this calculation:
t = (2 * Vy) / g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
t = (2 * 19.6) / 9.8
t ≈ 4 seconds
Now that we have the time, we can calculate the horizontal distance traveled by multiplying the horizontal velocity by the time:
distance = Vx * t
distance = 33.955 * 4
distance ≈ 135.82 meters
Therefore, the arrow will land approximately 135.82 meters out from the base of the cliff.