An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers. The plane is traveling horizontally at 40.0 m/s at a height of 100 m above the ground.
a) How far horizontally from the dropping point does the package strike the ground?
b)What are the x and y velocities before it hits the ground?
s = ut + 1/2 at^2.............since the initial velocity down u = 0
100 = 0 + 1/2(9.81)t^2
t = 4.51 sec
again s = ut + 1/2 at^2...........this time horizontally
s = distance = (40)(4.51) + 1/2(0)(4.51)^2................Notice the acceleration = 0
s = 180.61 m away in the direction of flight
this is the same as saying distance = speed x time.
II. Kinetic Energy = 1/2 mv^2
Potential Energy = mgh
(1/2)m(40)^2 + m(9.81)(100) = 1/2 m v^2............................
divide by m
(1/2) (1600) + 981 = 1/2(v^2)
v = 59.6 m/s
This ia according to the law of conservation of Energy.
Another way to do it is Horizontal and vertical velocity:
Horizontal v = 40 m/s...............since a = 0
Vertical v = u + at = 0 + (9.81)(4.51) = 44.23 m/s
Using pythagoras the resulting v = Squaroot ( 44.23^2 + 40^2 ) = 59.6 m/s
Same answer......and it should be, either way would do..
a) How far horizontally from the dropping point does the package strike the ground?
Well, since the plane was traveling horizontally at a constant speed, we can ignore any horizontal acceleration. This means that the horizontal distance the package travels is simply the horizontal velocity multiplied by the time it takes to reach the ground.
To find the time it takes to reach the ground, we can use the equation: h = (1/2)gt^2, where h is the height (100 m) and g is the acceleration due to gravity (9.8 m/s^2). Rearranging the equation to solve for t, we get:
t = sqrt(2h/g)
Plugging in the values, we find:
t = sqrt(2 * 100 / 9.8) ≈ 6.47 s
Now, to find the horizontal distance, we multiply the time by the horizontal velocity:
Distance = Velocity * Time
Distance = 40.0 m/s * 6.47 s
Distance ≈ 258.8 m
So, the package strikes the ground approximately 258.8 meters horizontally from the dropping point.
b) What are the x and y velocities before it hits the ground?
Since there is no horizontal acceleration, the x velocity remains constant at 40.0 m/s.
For the y velocity, we can use the equation: v = gt, where g is the acceleration due to gravity and t is the time of flight. Plugging the values, we have:
v = 9.8 m/s^2 * 6.47 s
v ≈ 63.6 m/s
So, the x velocity before it hits the ground is 40.0 m/s, and the y velocity is approximately 63.6 m/s.
To find the distance horizontally from the dropping point where the package strikes the ground (a), we can use the equation of motion:
d = V * t
where:
- d is the horizontal distance
- V is the horizontal velocity of the plane
- t is the time it takes for the package to hit the ground
To find the time, we need to determine how long it takes for the package to fall from a height of 100 m. We can use the kinematic equation:
h = (1/2) * g * t^2
where:
- h is the height
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time
Rearranging the equation to solve for t:
t = sqrt((2 * h) / g)
Plugging in the values:
t = sqrt((2 * 100) / 9.8)
t ≈ 4.52 seconds
Now we can find the horizontal distance:
d = V * t
d = 40.0 m/s * 4.52 s
d ≈ 180.8 meters
Therefore, the package strikes the ground approximately 180.8 meters horizontally from the dropping point (a).
To find the x and y velocities before the package hits the ground (b), we know that the horizontal velocity remains constant at 40.0 m/s. The vertical velocity can be determined using the equation:
v = g * t
Plugging in the values:
v = 9.8 m/s^2 * 4.52 s
v ≈ 44.4 m/s
Therefore, the x-velocity is 40.0 m/s and the y-velocity is approximately 44.4 m/s (b).
To find the answers to these questions, we can use the equations of motion. Let's break the problem down step by step:
a) How far horizontally from the dropping point does the package strike the ground?
To calculate the horizontal distance traveled by the package, we need to know the time it takes for the package to reach the ground. We can find this by using the equation of motion for vertical displacement:
s = ut + (1/2)gt^2
Where:
s = vertical displacement (in this case, the height from the plane to the ground)
u = initial vertical velocity (which is 0 since the package is dropped)
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values:
100 m = 0 + (1/2)(9.8 m/s^2)t^2
Simplifying the equation:
100 m = 4.9 m/s^2 * t^2
Now, solve for t:
t^2 = 100 m / 4.9 m/s^2
t^2 = 20.41 s^2
t ≈ 4.52 s
So, the package takes approximately 4.52 seconds to reach the ground.
To find the horizontal distance, we can use the equation:
d = vt
where:
d = horizontal distance (what we're trying to find)
v = horizontal velocity (40.0 m/s)
t = time taken (4.52 s, as we found earlier)
Plugging in the values:
d = 40.0 m/s * 4.52 s
d ≈ 180.8 m
Therefore, the package strikes the ground approximately 180.8 meters horizontally from the dropping point.
b) What are the x and y velocities before it hits the ground?
Since the package was dropped, there is no horizontal component of velocity, so the initial horizontal velocity (Vx) is 0 m/s.
The vertical velocity (Vy) can be found using the equation:
Vy = u + gt
Where:
u = initial vertical velocity (which is 0 since the package is dropped)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken (4.52 s)
Plugging in the values:
Vy = 0 + (9.8 m/s^2)(4.52 s)
Vy ≈ 44.3 m/s
Therefore, before hitting the ground, the package has a vertical velocity of approximately 44.3 m/s and a horizontal velocity of 0 m/s.