Answer:
3.83 km/s.
Explanation:
u = 95 m/s, h = 750 km = 750,000 m
Let the speed is v as it strikes the sand.
Use third equation of motion
v^2 = u^2 + 2 g h
v^2 = 95^2 + 2 x 9.8 x 750000
v^2 = 9025 + 14700000 = 14709025
v = 3835.23 m /s = 3.83 km /s
Thus, the speed of the meteorite as it strikes t sand is 3.83 km/s.
The speed of the meteorite just before striking the sand is 11,838.5 m/s
Vertical motion can be referred to motion under gravity.
Given that a meteorite has a speed of 95.0 m/s when 750 km above the Earth. It is falling vertically and strikes a bed of sand in which it is brought to rest in 3.35 m .
To know its speed just before striking the sand, we need to consider the radius of the earth R = 6400 km
The total distance S travelled = 750 + 6400
S = 7150 km
convert km to meters
S = 7150 x 1000 = 7150000 m
Let us assume that it is a free fall. The following parameters will be used.
Using 3rd equation of motion
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2gs
[tex]v^{2}[/tex] = [tex]95^{2}[/tex] + 2 x 9.8 x 7150000
[tex]v^{2}[/tex] = 9025 + 140140000
[tex]v^{2}[/tex] = 140149025
V = 11838.5 m/s
Therefore, its speed just before striking the sand is 11,838.5 m/s
Learn more about vertical motion here: https://brainly.com/question/24216590
