An Alaskan rescue plane traveling 49 m/s drops a package of emergency rations from a height of 177 m to a stranded party of explorers. The acceleration of gravity is 9.8 m/s 2 . Where does the package strike the ground relative to the point directly below where it was released?
Ok now the next part is, What is the horizontal component of the velocity just before it hits?
don't worry i figured the first part out, I just need the second one
There is NO horizontal force on the object
Therefore it moved at the forwards speed of the plane until it hits ground.
We already used that fact.
Remember:
"..... now horizontal problem for answer
goes at 49 m/s for t seconds from drop point ......"
To find the point where the package strikes the ground relative to the point directly below where it was released, we need to calculate the time it takes for the package to fall.
First, let's find the time it takes for the package to fall by using the kinematic equation:
d = v₀t + (1/2)at²
Where:
d = distance (177 m)
v₀ = initial velocity (0 m/s as the package was dropped)
a = acceleration due to gravity (-9.8 m/s^2, taking the negative sign as it opposes the motion)
t = time
Substituting the given values into the equation, we have:
177 = 0*t + (1/2)(-9.8)(t²)
Simplifying,
177 = -4.9t²
Rearranging the equation to isolate t², we have:
t² = 177/(-4.9)
t² ≈ -36.122
Since time cannot be negative, it means that there is an error in our calculations. However, we know that the package will not have a negative time of flight. This suggests that our initial assumption that the package was dropped with an initial velocity of 0 m/s is incorrect.
The problem states that the plane is traveling at 49 m/s. Therefore, we need to take into account the horizontal component of the plane's velocity.
Assuming the package is released directly below the point where it was dropped, we can ignore the horizontal component of the plane's velocity until the package hits the ground.
To find the time it takes for the package to fall, we can use the vertical component of the plane's velocity.
Using the equation:
d = v₀t + (1/2)at²
Since the initial vertical velocity(v₀) is 0 m/s, we have:
177 = 0*t + (1/2)(-9.8)(t²)
177 = -4.9t²
Simplifying further, we get:
t² = -177/(-4.9)
t² ≈ 36.122
Now we take the positive square root of t² to find t:
t ≈ √36.122
t ≈ 6.011 seconds
So, it takes approximately 6.011 seconds for the package to fall.
To find the horizontal distance traveled by the plane during this time, we can use the horizontal component of the plane's velocity:
distance = velocity * time
distance = 49 m/s * 6.011 s
distance ≈ 294.139 m
Therefore, the package strikes the ground approximately 294.139 meters away from the point directly below where it was released.
It proceeds under the plane at the same horizontal velocity component as the plane so it hits right under the plane (turn when you drop a bomb)
so 49 m/s until it hits ground
Vertical problem
v = -g t
z = 177 -(1/2) g t^2
at ground 0 = 177 -(1/2) 9.81 t^2
so
t^2 = 177/4.9
solve for t
now horizontal problem for answer
goes at 49 m/s for t seconds from drop point