Astudent throws a rock into the sky from the top of the webster build- ing with an angle α = π 4 from the horizontal line and the initial velocity 1 is 10 meters per second. the building is 100 meters tall from the ground. suppose there is only the gravitational force acting on it with the gravita- tional acceleration g = 9.8m/s 2 . suppose we choose the horizontal ground surface as x-axis and vertical line above the bullet position as y-axis. let r(t) =< x(t), y(t) > be the position function for the bullet at time t > 0. find the initial position r(0) =< xo, yo > and the initial velocity r'(0) =< uo, vo >

Step-by-step explanation:

Initial Position

The problem statement tells us we're measuring position from the ground at the base of the building where the projectile was launched. The initial horizontal position is presumed to be zero. The initial vertical position is said to be 100 meters from the ground, so (in meters) ...

  r(0) = <0, 100>

Initial Velocity

The velocity vector resolves into components in the horizontal direction and the vertical direction. For angle α from the horizontal, the horizontal component of velocity is v₁·cos(α), and the vertical component is v₁·sin(α). For v₁ = 10 m/s and α = π/4, the initial velocity vector (in m/s) is ...

  r'(0) = <10·cos(π/4), 10·sin(π/4)>

  r'(0) ≈ <7.071, 7.071>

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )