Answer:
42244138.951 m
Explanation:
G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²
r = Radius of orbit from center of earth
M = Mass of Earth = 5.98 × 10²⁴ kg
m = Mass of Satellite
The satellite revolves around the Earth at a constant speed
Speed = Distance / Time
The distance is the perimeter of the orbit

The Centripetal force of the satellite is balanced by the universal gravitational force

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m
The distance covered by an object accelerating from rest is
D = (1/2) · (acceleration) · (time)² .
In this particular case, 'acceleration' is 9.8 m/s² ... due to gravity.
D = (1/2) · (9.8 m/s²) · (1.67 s)²
D = (4.9 m/s²) · (2.789 s²)
D = 13.67 meters
Swimmer and Divers. The Potential energy is transferred into Kinetic energy, and allows the diver to submerge into the water. The Kinetic energy then allows the diver to submerge and dive into the water. Potential energy however, is needed to allow the diver to get back out of the water after diving to get up and go and dive again, and then the Kinetic energy is transferred back to Potential energy to repeat the process.
Hope :) -Emilie Xo this is right and it helps! Xo
Answer:
t = 3.48 s
Explanation:
The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

where,
v₀ = initial speed = 110 ft/s
Therefore,

<u>t = 3.48 s</u>
Answer:
x = 17.88[m]
Explanation:
We can find the components of the initial velocity:
![(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7Bo%7D%20%20%3D%2013.3%2Acos%2841.5%29%3D9.96%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7Bo%7D%20%20%3D%2013.3%2Asin%2841.5%29%3D8.81%5Bm%2Fs%5D)
We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.
g = - 9.81[m/s^2]
Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.
![y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2B%28v_%7By%7D%20%29_%7Bo%7D%20%2At-0.5%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5C0%3D1.9%2B%288.81%2At%29-%284.905%2At%5E%7B2%7D%29%5C%5C-1.9%3D8.81%2At%2A%281-0.5567%2At%29%5C%5Ct%3D0%5C%5Ct%3D1.796%5Bs%5D)
With this time we can calculate the horizontal distance:
![x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bx%7D%29_%7Bo%7D%20%2At%5C%5Cx%3D9.96%2A1.796%5C%5Cx%3D17.88%5Bm%5D)