A rescue plane flying horizontally at 72.6 m / s spots a survivor in the ocean 182 m directly below and releases an emergency kit with a parachute. Because of the shape of the parachute, it experiences insignificant horizontal air resistance. If the kit descends with a constant vertical acceleration of 5.82 m / s 2 , how far away from the survivor will it hit the waves?

547 m

Explanation:

From law of motion

s = ut + ½at²

Where "t" is Time taken to reach Earth

s= distance= 182 m

a= vertical acceleration = 5.82 m / s 2

U= initial velocity in vertical position = 0

182= ½ × 5.82t²

t²=( 2× 182)/ 5.82

= 364/5.82

= 62.54

t= √62.54

t= 7.908s

horizontal distance travelled = speed x time

Horizontal speed= 72.6 m / s

horizontal distance travelled =72.6× 7.908

= 547 m

Hence, the survivor will it hit the waves at 547 m away

im not to sure what it is

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.