A plane is flying horizontally with speed 162 m/s at a height 3970 m above the ground,
when a package is dropped from the plane.
2
Neglecting air resistance, when the package hits the ground, the plane will be
1. directly above the package. 2. ahead of the package.
3. behind the package.

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.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.