A military supply plane is on a mission to drop supplies. Its flying 1200 m above the ground at 723 m/s to avoid enemy fire to drop its pay load of (800 kg) on the correct target. Given The mass of the 800kg of the supply run lands on a frozen lake. It travels for 30 sec on the lake before hitting the snow.
a)How far did it travel?
b)After 30 sec, it hits snow (which has a coefficient of friction of 0.2), how far does it travel in the snow?
c) How far does it travel horizontally in the air,, frozen lake, and snow together?
Show work please

a. 21690 m. b. 20808 m c. 53812.95 m

Explanation:

a. Since the horizontal velocity of the pay load equals that of the military plane, v = 723 m/s. The distance it moves on the frozen lake in 30 s is d = vt = 723 m/s × 30 s = 21690 m.

b. Since the only force acting of the payload when it hits the snow is friction and is opposite to its direction of motion. Then

-f = ma and f = μmg where f = frictional force, μ = coefficient of friction of snow = 0.2, m = mass of payload = 800 kg ,a = acceleration of payload and g = 9.8 m/s²

-μmg = ma

a = -μg = -0.2 × 9.8 m/s² = -1.96 m/s²

We then use s = ut + 1/2at² to find the distance moved after 30 s. u = 723 m/s, a = -1.96 m/s² and t = 30 s

s = ut + 1/2at²

= 723 m/s × 30 + 1/2 × (-1.96 m/s²) ×30²

= 21690 m - 882 m

= 20808 m

c. Since the payload is falling under gravity, we calculate the time it takes to hit the frozen lake which is also equal to the time it stays in the air from

s = ut + 1/2at². s = 1200 m, u = initial vertical velocity = 0 m/s, a = g = -9.8 m/s²

-1200 m = 0t + 1/2 × -9.8 m/s² × t².

-1200  = 0 - 4.9t².

1200  = 4.9t².

t = √1200/4.9

= 15.65 s

The horizontal distance d the payload moves in the air is thus d = vt = 723 m/s × 15.65 s = 11314.95 m

The total horizontal distance is thus 21690 m + 20808 m + 11314.95 m = 53812.95 m

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures