Answer:
2. ahead of the package.
Explanation:
Using y' - y = ut - 1/2gt², we find the time, t it takes the package to hit the ground. So, u = initial vertical velocity of package = 0 m/s, y = initial position of package = 3970 m, y' = final position of package = 0 m, g = acceleration due to gravity = 9.8 m/s².
Substituting the variables into the equation, we have
y' - y = ut - 1/2gt²
0 m - 3970 m = 0t - 1/2 × (9.8 m/s²)t²
-3970 m = -(4.9m/s²)t²
t² = -3970 m ÷ -4.9 m/s²
t² = 810.2 s²
t = √810.2 s²
t = 28.5 s
Using v = u' + at, we find the horizontal acceleration of the plane. Since the initial horizontal velocity of the package is that of the plane, u' = 162 m/s, v = 0 m/s since the package stops and t = 28.5 s when the package stops.
So, a = (v - u')/t
a = (0 m/s - 162 m/s)/28.5 s
a = -162 m/s/28.5 s
a = -5.68 m/s²
Using v² = u'² + 2as, we find the horizontal distance ,s where the package stops.
So, s = (v² - u'²)/2a
Substituting the values of the variables, we have
s = ((0 m/s)² - (162 m/s)²)/(2 × -5.68 m/s²)
= - 162 m²/s²/(-11.36 m/s²)
= 14.26 m
The horizontal distance d the plane moves after releasing the package is d = u't = 162 m/s × 28.5 s = 4617 m
Since d = 4617 m > s = 14.26 m, the plane would be ahead of the package when the package hits the ground.
