A rescue plane drops a package of emergency rations to a stranded explorer. The plane is travelling horizontally at 40 m/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 equations to solve this problem. The key equation we need is:
y = vi*t + 1/2*a*t^2
where y is the vertical displacement (i.e. the height above the ground), vi is the initial velocity (which is zero when the package is released), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time elapsed.
We can use this equation to find the time it takes for the package to hit the ground:
100 = 0 + 1/2*(-9.8)*t^2
t = sqrt(2*100/9.8) = 4.52 s
Now we can use the horizontal velocity of the plane (40 m/s) and the time elapsed (4.52 s) to find the horizontal distance travelled by the package:
x = vi*t + 1/2*a*t^2
x = 40*4.52 = 181.04 m
So the package hits the ground 181.04 m horizontally away from the point it was released.
To find out where the package hits the ground relative to the point it was released, we can analyze the horizontal and vertical motions separately.
First, let's consider the horizontal motion. Since the plane is traveling horizontally at a constant speed of 40 m/s, the horizontal distance the package travels is solely determined by the time of flight. We can use the equation:
Distance = Speed x Time
In this case, the speed is 40 m/s, and let's assume the time of flight is denoted as "t." Therefore, the horizontal distance traveled by the package is:
Horizontal Distance = 40 m/s x t
Now, let's focus on the vertical motion of the package. We'll use the equation for the vertical displacement:
Vertical Displacement = (Initial Velocity x Time) + (0.5 x Acceleration x Time^2)
In this case, the initial velocity of the package is 0 m/s since it's just dropped from the rescue plane. The acceleration due to gravity, denoted as "g," is approximately 9.8 m/s^2. Again, the time of flight is denoted as "t."
As the package falls from a height of 100 m above the ground, the vertical displacement becomes:
Vertical Displacement = (0 m/s x t) + (0.5 x 9.8 m/s^2 x t^2)
Vertical Displacement = 4.9 m/s^2 x t^2
Since the package falls vertically, the vertical displacement is also equal to the height from which it was dropped. Therefore:
Vertical Displacement = -100 m
Combining the equations for horizontal and vertical motion, we find that both motions occur at the same time (t). Thus, we can equate the horizontal and vertical displacements:
40 m/s x t = -100 m
Solving for "t," we get:
t = -100 m / 40 m/s
t = -2.5 s
Since time cannot be negative in this scenario, it means that the package is in the air for 2.5 seconds.
Now, we can find the horizontal distance traveled by substituting this value of "t" into the equation:
Horizontal Distance = 40 m/s x 2.5 s
Horizontal Distance = 100 m
Therefore, the package hit the ground 100 meters horizontally from the point it was released.