Ahelicopter carrying supplies is traveling eastward, 80.0 m off the ground, at a speed of 40.0 m/s. the supplies are to be dropped from the helicopter to land in the back of a parked pickup truck on a flat road. when the helicopter and the truck are 300 m apart,
(a) how many seconds later should the helicopter crew drop the supplies?
(b) if the pickup truck was traveling westward at 25.0 mi/h from the beginning, then at what time should the helicopter crew drop the supplies?

a) Helicopter need to drop the supply after 3.46 seconds.

b) Helicopter need to drop the supply after 1.83 seconds.

Explanation:

a) Let us find time of flight of supplies that is vertical motion of supplies,

We have equation of motion s = ut + 0.5at²

          Initial velocity, u = 0 m/s

         Acceleration, a = 9.81 m/s²

         Displacement, s = 80 m

    Substituting,

                  s = ut + 0.5 at²

                 80 = 0 x t + 0.5 x 9.81 x t²

                    t = 4.04 s

  Distance traveled by helicopter in 4.04 seconds at 40 m/s = 4.04 x 40 = 161.54 m

  Distance helicopter can fly without dropping supply = 300 - 161.54 = 138.46 m

 Time taken for this

   So helicopter need to drop the supply after 3.46 seconds.

b) Speed of truck = 25 mi/h = 11.11 m/s westward

   Speed of helicopter = 40 m/s eastward

   Relative velocity, = 40 + 11.11 = 51.11 m/s

  We have equation of motion s = ut + 0.5at²

          Initial velocity, u = 0 m/s

         Acceleration, a = 9.81 m/s²

         Displacement, s = 80 m

    Substituting,

                  s = ut + 0.5 at²

                 80 = 0 x t + 0.5 x 9.81 x t²

                    t = 4.04 s

  Distance traveled by relative bodies  in 4.04 seconds at 51.11 m/s = 4.04 x 51.11 = 206.48 m

  Distance helicopter can fly without dropping supply = 300 - 206.48 = 93.52 m

 Time taken for this

   So helicopter need to drop the supply after 1.83 seconds.  

real, smaller and upside-down.   so the answer should be b.

no sign of dia of fuel cell