this can be solve using the formala of free fall
t = sqrt( 2y/ g)
where t is the time of fall
y is the height
g is the acceleration due to gravity
48.4 s = sqrt (2 (1.10e+02 m)/ g)
G = 0.0930 m/s2
The velocity at impact
V = sqrt(2gy)
= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)
V = 4.523 m/s
<span> </span>
a) KE=0.5*mv^2==0.5*145*25^2=45312.5 J
b) PE=mgh=145*9.8*3.5=4973.5 J
c) ME=KE+PE=m(0.5v^2+gh)=62524 J
1. earth
2. solar system
3. milky way galaxy
4. local group of galaxies
5. universe
Answer:
a) N = 343 [N]
Explanation:
We must remember Newton's third law, which tells us that the force acting on a body is equal to the normal reaction force of equal magnitude but acting in the opposite direction.

where:
m = mass = 35 [kg]
g = gravity acceleration = 9.81 [m/s²]
![N = 35*9.8\\N= 343 [N]](https://tex.z-dn.net/?f=N%20%3D%2035%2A9.8%5C%5CN%3D%20343%20%5BN%5D)