Answer:
333 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 0 m/s.
Time (t) = 8.14 s.
Final velocity (v) = 80.79 m/s.
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
v² = u² + 2gh
80.79² = 0 + (2 × 9.8 × h)
6527.0241 = 0 + 19.6h
6527.0241 = 19.6h
Divide both side by 19.6
h = 6527.0241 / 19.6
h = 333 m
Thus, the plane was at a height of 333 m when the package was dropped.
Displacement is the area under the velocity/time graph. So for example this object's displacement in the first 3 seconds is (1/2)(3sec)(12.5 m/s)= 18.75m. (and then it starts backing up, displacement decreasing, after 3sec when velocity is negative).
But This object is never speeding up. Its velocity is smoothly decreasing at (25/6) m/s^2 (the slope of the graph). So the answer to the question is actually zero.
a/b. The ball has velocity vector at time 

where
is the ball's initial speed and
.
c. At its highest point, the ball has 0 vertical speed. This occurs when

d. Recall that

so that at its highest point,

e. This is just twice the time it takes for the ball to reach its maximum height,
.
f. The ball's horizontal position after time
is

so that after the time found in part (f), the ball has traveled
