A food package was dropped from an aircraft flying horizontally. 6 seconds before it hit the ground, it was at a height of 780 m, and had traveled a distance of 1 km horizontally. Find the speed and the altitude of the aircraft.

Now, when the aircraft will drop the food package, it will obviously travel a parabolic path. Let the aircraft be flying with the horizontal speed = v m·s−1. So, the horizontal velocity given to the food package when left will be same = v m·s−1. The vertical component will be zero i.e. it is coming down by the effect of gravity. Let us say the vertical height (altitude) of the aircraft = x m. Let us say the total time taken by the package to reach the ground = t seconds. So, Initial vertical downward velocity u = 0 m·s−1. Distance = -x  ( Vertical Downward displacement) Acceleration due to gravity (g) = -10 m·s−2 Time = t seconds So -x = 1/2 (-10) t² Cancelling the negative sign, x = 5 *  t² Now in last 6 seconds, package covered the distance of 780 m So, distance covered in (t-6) seconds = (x - 780) m Now, again applying the equation-(x-780) = 1/2 (-10) (t-6)² Cancelling the negative sign, and solving we get  t = 16 seconds Putting in first equation x = 5 * t² we get x = 1280 m = 1.28 km (Answer) Since, it has travelled a horizontal distance of 1000 m before 6 seconds, so this distance has been covered in (16 - 6 = 10 s) Distance = Speed * Time 1000 = v * 10 (Horizontal component given to food package is same as speed of aircraft) v = 100 m·s−1 = 360 km/ hr (Answer).