Question

A plane is flying horizontally with speed 292 m/s at a height 3880 m above the ground, when a package is dropped from the plane. the acceleration of gravity is 9.8 m/s 2 . neglecting air resistance, when the package hits the ground, the plane will be 1. ahead of the package. 2. directly above the package. 3. behind the package. 008 (part 2 of 4) 10.0 points what is the horizontal distance from the release point to the impact point?

Answer 1

The horizontal distance from the release point to the impact point is approximately 8446.24 meters.

Step-by-step explanation:

To find the horizontal distance from the release point to the impact point, we need to use the equation of motion for horizontal motion:

d = vt

In this equation, d is the horizontal distance, v is the horizontal velocity, and t is the time of flight. The time of flight can be found using the equation:

t = sqrt(2h/g)

In this equation, h is the initial height and g is the acceleration due to gravity. Plugging in the given values, we have:

t = sqrt(2 * 3880 / 9.8) = 28.92 s

Substituting this value into the equation for horizontal distance, we have:

d = (292 m/s) * (28.92 s) = 8446.24 m

Therefore, the horizontal distance from the release point to the impact point is approximately 8446.24 meters.

Answer 2

The plane is straight above the package when it hits the ground.
Find in what way long a free fall from a height of 4240 m takes: s = 1/2 gt^2
t^2 = 2s/g
t^2 = 2* 3880/9.8 = 791.8367
t = √791.8367
t = 28.1396 seconds
Now find the horizontal displacement with v = 292 m/s in 28.1396 s:
s = v*t
s = 292 * 28.1396 = 8216.76 m