From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 9.40 m/s and angle of 23.0° below the horizontal. It strikes the ground 6.00 s later.

Initial velocity components:
vx = 8.65 m/s
vy = -3.67 m/s

(c) Find the equations for the x- and y- components of the position as functions of time.
x = 8.65t
y = ?

(d) How far horizontally from the base of the building does the ball strike the ground?
51.9 m

(e) Find the height from which the ball was thrown.

(f) How long does it take the ball to reach a point 10.0 m below the level of launching?

I need help on (c), (e), and (f).

Vo = 9.4m/s[23o]

Xo = 9.4*cos23 = 8.65 m/s.
Yo = 9.4*sin23 = 3.67 m/s.

c. X = Xo * t = 8.65t
Y = Yo*t + 0.5g*t^2=3.67t+4.9t^2

e. h = 3.67t + 4.9t^2
h = Y = 3.67*6 + 4.9*6^2 = 198.4 m.
Above gnd.

f. 3.67t + 4.9t^2 = 10 m.
4.9t^2 + 3,67t _- 10 = 0
t = 1.10 s. (Use Quad. Formula).

To solve part (c), we need to find the equation for the y-component of the position as a function of time.

The y-component of the position can be found using the equation:
y = y0 + vy*t + (1/2)*a*t^2

As the ball is thrown horizontally, the acceleration in the y-direction is due to gravity, and its magnitude is 9.8 m/s^2. We'll assume the positive y-direction is upwards and the negative y-direction is downwards.

Given that the initial height above the ground (y0) is y0, the initial vertical velocity (vy) is -3.67 m/s, and the time (t) is 6.00 s, we can substitute these values into the equation:

y = y0 + vy*t + (1/2)*a*t^2
y = y0 + (-3.67 m/s)*(6.00 s) + (1/2)*(9.8 m/s^2)*(6.00 s)^2

You'll need to know the initial height of the ball (y0) to calculate the vertical position of the ball as a function of time.

For part (e), we are asked to find the height from which the ball was thrown. To determine this, we need to calculate y0.

Given that the ball is thrown from a building, the initial height above the ground can be represented by y0. This value is not provided in the given information, so you'll need to obtain it from the problem statement or any additional information given.

Finally, for part (f), we are asked to find the time it takes the ball to reach a point 10.0 m below the level of launching.

To solve for the time, we can use the equation for the y-component of the position as a function of time and set the position (y) equal to the desired height (10.0 m). Then, solve for time (t):

10.0 = y0 + vy*t + (1/2)*a*t^2

Substituting the known values for vy, a, and y0 into the equation allows you to solve for time (t).