An airplane is flying with a velocity of 95.0 at an angle of 26.0 above the horizontal. When the plane is a distance 113 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.?
assuming the unit of distance here is meters,
the horizontal speed is 95 cos26° = 85.4m/s
to get time to fall:
113 = 4.9t^2
t = 4.8 seconds
horizontal distance moved is thus
85.4 * 4.8 = 410.1 m
assuming the dog doesn't start chasing the plane! :-)
Obituary
To find the horizontal distance from the dog to the point where the suitcase will land, we can use the horizontal component of the airplane's velocity.
First, let's find the horizontal component of the airplane's velocity using trigonometry:
horizontal velocity = velocity * cos(angle)
horizontal velocity = 95.0 * cos(26.0)
horizontal velocity ≈ 86.2
Now, let's find the time it takes for the suitcase to reach the ground. We can use the formula:
time = sqrt((2 * vertical distance) / acceleration)
Since the suitcase falls vertically, the vertical distance is equal to the height from the dog to the luggage compartment, which is 113 meters.
acceleration due to gravity, assuming no air resistance, is approximately 9.8 m/s^2.
time = sqrt((2 * 113) / 9.8)
time ≈ sqrt(23 / 9.8)
time ≈ sqrt(2.35)
time ≈ 1.53 seconds
Finally, we can find the horizontal distance using the formula:
horizontal distance = horizontal velocity * time
horizontal distance = 86.2 * 1.53
horizontal distance ≈ 132.29
Therefore, the suitcase will land approximately 132.29 meters away from the dog.
To find the horizontal distance from the dog where the suitcase will land, we need to determine how long it takes for the suitcase to hit the ground and then calculate the horizontal distance traveled during that time.
Step 1: Determine the vertical motion of the suitcase
Since the suitcase is dropped from the airplane, its initial vertical velocity is 0 m/s. Therefore, we can use the equation for vertical motion:
h = ut + (1/2)gt^2,
where h is the initial vertical displacement (in this case, 113 m), u is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Substituting these values into the equation, we get:
113 = 0*t + (1/2)(9.8)*t^2.
Simplifying the equation, we have:
113 = 4.9t^2.
Rearranging the equation to solve for t, we get:
t^2 = 113 / 4.9.
t^2 = 23.061.
t ≈ 4.8 seconds.
Step 2: Determine the horizontal distance traveled by the suitcase
To find the horizontal distance traveled by the suitcase, we can use the equation for horizontal motion:
d = vt,
where d is the horizontal distance, v is the horizontal velocity, and t is the time.
The horizontal velocity can be found using the given velocity and angle. Considering only the horizontal component of the velocity, we have:
vx = velocity * cos(angle).
Substituting the given values into the equation, we get:
vx = 95 * cos(26).
vx ≈ 85.171 m/s.
Now, we can calculate the horizontal distance traveled by the suitcase:
d = 85.171 * 4.8.
d ≈ 409.939 m.
Therefore, the suitcase will land approximately 409.939 m from the dog.