Question

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

Answer 1

A.) 156.5 mts. B) directly over target.

Step-by-step explanation:

A.) Once dropped, the object is under the only influence of gravity, which acts always downward.

So, in the horizontal direction, as no external force is acting, it must continue moving at the same speed that it had when it was dropped, which it was the same as the plane.

So, we can write two kinematic equations, one for the horizontal movement and the other for the vertical one, as follows:

1) Horizontal:

Δx = vx*t

2) Vertical:

h = 1/2 gt² (as the vertical initial speed is 0, due to the weight is dropped)

We can solve for time this equation, as follows:

t² = 2*h/g ⇒t=√(2*h/g) = √2*75m/9.8 m/s² = 3.9 sec

As the time, obviously is the same for both directions, we can replace this value in the equation for the horizontal distance, as follows:

Δx = vx*t = 40 m/s*3.9 sec = 156.5 m

B) Assuming no change in speed in the instant that the weight was dropped, the plane continued moving at the same speed that it had, which is the same that the weight had in the horizontal direction, so the plane is always directly above the weight along all the trajectory, including the instant in which the weight hits the ground.

So, at that instant, the plane is flying directly over the target.