A rescue plane drops a package of emergency rations to a stranded explorer. The plane is travelling at 40m/s and at a height of 100m above the ground. Where does the package hit the ground relative to the point it was released?
We can use the kinematic equation:
y = yi + vi*t + (1/2)*a*t^2
where
y = final height (relative to the point of release)
yi = initial height (100m above the point of release)
vi = initial velocity (0 m/s since the package is dropped from rest)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach the ground
Since the package is dropped, we can assume that its initial velocity is 0 m/s. We can rearrange the equation to solve for t:
t = sqrt((2*yi)/a)
t = sqrt((2*100)/9.8) = 4.04 s
Now we can use another kinematic equation to find the horizontal distance travelled by the package:
x = xi + vi*t
where
x = horizontal distance travelled
xi = initial horizontal position (unknown)
vi = horizontal velocity (40 m/s)
t = time taken (4.04 s)
x = vi*t = (40 m/s)*4.04 s = 161.6 m
Therefore, the package hits the ground 161.6 m horizontally away from the point of release.
To determine where the package hits the ground relative to the point it was released, we can use the kinematic equations of motion. First, let's find the package's time of flight (t).
We can use the following equation to find the time of flight:
h = ut + (1/2)gt^2
Where:
h = initial vertical displacement (100m)
u = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time of flight
Plugging in the values, the equation becomes:
100 = 0*t + (1/2)*(-9.8)*t^2
Simplifying and rearranging the equation, we have:
4.9t^2 = 100
Dividing both sides by 4.9, we get:
t^2 = 20.41
Taking the square root of both sides, we find:
t ≈ 4.52 seconds
The package takes approximately 4.52 seconds to reach the ground.
Next, we can find the horizontal distance traveled by the package during this time. We can use the equation:
s = ut
Where:
s = horizontal distance traveled
u = initial horizontal velocity (40m/s)
t = time of flight (4.52 seconds)
Plugging in the values, we have:
s = 40 * 4.52
s ≈ 180.8 meters
Therefore, the package hits the ground approximately 180.8 meters horizontally from the point it was released.