A rescue plane drops a package of emergency supplies 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 solve this problem using kinematic equations. Let's assume that the package is dropped from rest:
Initial velocity of package (u) = 0 m/s
Acceleration due to gravity (a) = 9.81 m/s^2 (assuming no air resistance)
Height of package above ground (h) = 100 m
Using the equation of motion, h = ut + 1/2(at^2), we can find the time taken for the package to hit the ground:
100 = 0t + 1/2(9.81t^2)
Simplifying,
4.905t^2 = 100
t^2 = 20.39
t = 4.51 seconds (rounded to 2 decimal places)
Now, using the equation of motion, s = ut + 1/2(at^2), we can find the horizontal distance travelled by the package during this time:
s = 40 x 4.51 + 1/2(0)(4.51)^2
s = 180.4 meters
Therefore, the package hits the ground 180.4 meters away from the point it was released, in the horizontal direction.
To determine where the package lands relative to the point it was released, we can use the formula for the horizontal distance traveled (x) by an object in projectile motion:
x = v * t
where:
- v is the horizontal velocity of the plane (40 m/s)
- t is the time it takes for the package to reach the ground
To find the time (t), we need to consider the vertical motion of the package. We can use the equation for the vertical displacement (y) of an object in free fall:
y = ut + (1/2) * a * t^2
where:
- u is the initial vertical velocity of the package (0 m/s, since the package is dropped)
- a is the acceleration due to gravity (-9.8 m/s^2, as it acts downwards)
- t is the time it takes for the package to reach the ground
Since the package starts at a height of 100m above the ground, its initial vertical displacement (y) is 100m. At the ground level, its final vertical displacement is 0 m. Substituting these values into the equation, we have:
0 = (0 * t) + (1/2) * (-9.8) * t^2
0 = -4.9 * t^2
Solving for t, we find that t = 0 or t = √(0/(-4.9)), which simplifies to t = 0 or t = 10 s.
Since the time cannot be negative, we discard t = 0. Therefore, it takes 10 seconds for the package to reach the ground.
Now, using the value of t in the equation for horizontal distance traveled, we can calculate x:
x = v * t
x = 40 * 10
x = 400 m
Therefore, the package hits the ground 400 meters horizontally away from the point it was released.