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?

Answer 1

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.