Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ignored) from their low-flying planes and scoring points for having the weights land close to a target. a plane 75m above the ground is flying directly toward a target at 40m/s .
a) at what distance from the target should the pilot drop the weight?
b) the pilot looks down at the weight after she drops it. where is the plane located at the instant the weight hits the ground?
not yet over the target
past the target
directly over the target
not enough information to determine

a. 156.41m

b. directly over the target

Explanation:

a) At the point of release, the weight will have:

Horizontal:

velocity of 40m/s (same as the plane)

Acceleration of zero since it is not under the influence of any force.

Vertical:

initial Velocity - zero at this instant

Acceleration - Towards the ground with g (9.81m/s^2)

from newton's equation of motion, since its falling under gravity, the horizontal component wont be involved

s = ut + (1/2)at^2

u = initial velocity

a = acceleration

s = displacement

t = time taken

75 = 0 + (1/2)(9.81)t^2

t^2=15.29

t=3.91

t = 3.91s

Applying the time of flight to the horizontal components:

Distance = speed x time

= 40 x 3.91

= 156.41m

B) The pilot looks down at the weight after she drops it. Where is the plane located at the instant the weight hits the ground?

not yet over the target

past the target

directly over the target

not enough information to determine

c. directly over the target. will be the correct answer