Answer:
The distance will be x = 41.7 [m]
Explanation:
We must first find the components in the x & y axes of the initial velocity.
![(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]](https://tex.z-dn.net/?f=%28v_%7Bo%7D%29_%7Bx%7D%20%3D%2015%2Acos%2820%29%3D%2014.09%5Bm%2Fs%5D%5C%5C%28v_%7Bo%7D%29_%7By%7D%20%3D%2015%2Asin%2820%29%3D%205.13%5Bm%2Fs%5D)
The acceleration is the gravity acceleration therefore.
g = 9.81 [m/s^2]
Now we can calculate how long it takes to fall.
![y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]](https://tex.z-dn.net/?f=y%3D%28v_%7Bo%7D%29_%7By%7D%2At-0.5%2Ag%2At%5E2%5C%5C-28%20%3D%205.13%2At-0.5%2A9.81%2At%5E2%5C%5C-28%3D-4.905%2At%5E2%2B5.13%2At%5C%5C4.905%2At%5E2-5.13%2At%3D28%5C%5Ct%20%3D%202.96%5Bs%5D)
With this time we can find the horizontal distance that runs the projectile.
![x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bo%7D%29_%7Bx%7D%2At%5C%5Cx%3D14.09%2A2.96%5C%5Cx%3D41.7%5Bm%5D)
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and 
The sum is represented as
For the the values given to us the sum is calculated as

Now the since the uncertainity inthe sum is 
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals
meters
The object with the mass ok 1kg will move more quickly because it is lighter than the 100kg object
Answer:
Her angular speed (in rev/s) when her arms and one leg open outward is 
Explanation:
Initial moment of inertia when arms and legs in is 
Final moment of inertia when her arms and on leg open outward, 
Initial angular speed 
Let the final angular speed be 
Since external torque on her is zero so we can apply conservation of angular momentum

=>
=>
Thus her angular speed (in rev/s) when her arms and one leg open outward is 