At some speed, the drag or force of resistance will equal the gravitational pull on the object. At this point the object ceases to accelerate and continues falling at a constant speed called the terminal velocity (also called settling velocity).
Answer:
104.3 cm or 179.7
Explanation:
First find time that it takes for the object to hit the ground
*
Then find xf of projectile 
not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7 or 142+37.7
hope that helps
Impulse: a certain amount of force you apply for an amount of time.
Impulse: F*t where F= Force & t=time
Momentum: increasing forward motion.
A ball rolling down a slide gains momentum
p=mv where m=mass and v=velocity
Hope it helps!
~Just an emotional teen who listens to music
B) All nonzero digits are significant.
Answer: 
Explanation:
This problem can be solved using the Third Kepler’s Law of Planetary motion:
<em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
<em />
This law states a relation between the orbital period 
of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size 
of its orbit:
(1)
Where:
is the period of the orbit of the exoplanet (considering 
)
is the Gravitational Constant and its value is 
is the mass of the star
is orbital radius of the orbit the exoplanet describes around its star.
Now, if we want to find the radius, we have to rewrite (1) as:
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%5E%7B2%7D%7D%7D)
(2)
![a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%28122044320s%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%283.59%2810%29%5E%7B30%7Dkg%29%7D%7B4%5Cpi%5E%7B2%7D%7D%7D)
(3)
Finally:
This is the radius of the exoplanet's orbit
