An airplane is flying with a velocity of 86.0 m/s at an angle of 20.0^\circ above the horizontal. When the plane is a distance 119 m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.How far from the dog will the suitcase land? You can ignore air resistance.
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To find the horizontal distance at which the suitcase will land, we first need to find the time it takes for the suitcase to reach the ground. We can use the vertical motion equation:
y = y0 + v0yt - (1/2)gt^2
where:
y = final vertical position (0, since the suitcase is on the ground)
y0 = initial vertical position (119 m)
v0y = vertical component of the initial velocity (v0y = v0 * sin(θ))
g = acceleration due to gravity (9.8 m/s^2)
t = time
Since the final vertical position is 0, the equation becomes:
0 = 119 + (v0 * sin(θ))t - (1/2)gt^2
Rearranging the equation, we get:
(1/2)gt^2 - (v0 * sin(θ))t - 119 = 0
Now we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / 2a
where:
a = (1/2)g
b = -(v0 * sin(θ))
c = -119
Substituting the values, we get:
t = (-(v0 * sin(θ)) ± √((v0 * sin(θ))^2 - 4(1/2)g(-119))) / 2(1/2)g
Simplifying further:
t = (-(v0 * sin(θ)) ± √((v0 * sin(θ))^2 + 2g(119))) / g
Now we can plug in the given values:
v0 = 86.0 m/s
θ = 20.0°
g = 9.8 m/s^2
Calculating:
t = (-(86.0 * sin(20.0°)) ± √((86.0 * sin(20.0°))^2 + 2(9.8)(119))) / 9.8
Calculating the positive value, we get:
t = 5.32 seconds
Now we can find the horizontal distance traveled by the suitcase using the formula:
x = v0x * t
where:
x = horizontal distance
v0x = horizontal component of the initial velocity (v0x = v0 * cos(θ))
Using the given values:
v0 = 86.0 m/s
θ = 20.0°
Calculating:
x = 86.0 * cos(20.0°) * 5.32
The horizontal distance from the dog where the suitcase will land is approximately 439.22 m.