An airplane needs to drop a package at a specified drop site. The airplane cruises 5.00 x 10^3 meters above the ground at 350.0 km/hr. Assuming that there is no wind blowing and that the plane is flying in a direction that will take it exactly over the drop site, at what horizontal distance in front of the drop site should the package be released?
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To determine the horizontal distance in front of the drop site where the package should be released, we can use the equation relating distance, time, and velocity.
Given:
Height of the plane above the ground (h) = 5.00 x 10^3 meters
Speed of the plane (v) = 350.0 km/hr
First, we need to convert the speed of the plane from km/hr to m/s, since the height of the plane is given in meters.
1 kilometer = 1000 meters
1 hour = 3600 seconds
So, the speed of the plane in m/s is:
350.0 km/hr * (1000 m/1 km) * (1 hr/3600 s) = (350.0 * 1000) / 3600 ≈ 97.222 m/s
Now, we can calculate the time taken (t) for the package to reach the ground using the height of the plane and the acceleration due to gravity.
The equation for calculating time is:
h = (1/2) * g * t^2
where:
h is the height,
g is the acceleration due to gravity (approximately 9.8 m/s²),
and t is the time.
Rearranging the equation to solve for t, we get:
t = sqrt((2h / g))
Plugging in the values into the equation:
t = sqrt((2 * 5.00 x 10^3) / 9.8) ≈ sqrt(1020.41) ≈ 31.95 s
Now, using the time and the horizontal velocity of the plane, we can calculate the horizontal distance (d).
The equation for calculating distance is:
d = v * t
where:
d is the distance,
v is the velocity,
and t is the time.
Plugging in the values into the equation:
d = 97.222 m/s * 31.95 s ≈ 3107.66 meters
Therefore, the package should be released approximately 3107.66 meters in front of the drop site.