Please help me answer this question. I have finished the other two parametric equations on my own, but I am confused as to how to do this one.

4. A ball is hit at an angle of 17¢X. The ball is hit when it is 2.5 feet above the ground and is hit at a velocity of 136 feet per second.

Using the following parametric equations for the projectile motion problem.

x=(v0cos£c)t
y=h+(v0sin£c)t-16t^2

where v0 represents the initial velocity of the object (in feet per second), and £c represents that angle made with the horizontal. The value of h is the initial height of the object.
a. Write the two parametric equations that represent this situation.
b. At what time will the ball hit the ground? Round your answer to the nearest thousandth AND explain/show how you got your answer.
c. How far has the ball traveled when it hits the ground? Round to the nearest tenth of a foot AND show/explain how you got your answer.
d. If a 10 foot fence is positioned 327 feet from where the ball is hit, will the ball make it over the fence? If so, by how many feet does it clear the fence? If not, how far from the top of the fence does the ball hit?

Vo = 136Ft/s[17].

Xo = 136*cos17 = 130 Ft/s.
Yo = 136*sin17 = 39.8 Ft/s.

b. Y = Yo + g*Tr = 0, 39.8 - 32Tr = 0, Tr = 1.244 s. = Rise time.

h = 2.5 + 39.8*1.244 - 16*1.244^2 = 27.25 Ft a5ove gnd.

16*Tf^2 = 27.25, Tf = 1.305 s. = Fall time.

1.244+1.305 = 2.549 s. to hit ground.

c. Dx = Xo*(Tr+Tf) = 331.37 Ft.

d. 16*Tf^2 = 27.25-10 = 17.25.

Tf = 1.04 s.

Dx = Xo*(Tr+Tf) = 130*2.28 = 296.7 Ft.

327-296.7 = 30.3 Ft. from bottom of fence.

d^2 = 30.3^2 + 10^2 = 1019.3, d = 31.9 Ft from top of fence.

a. To represent this situation using parametric equations, we'll substitute the given values into the equations provided:

Given:
v0 = 136 ft/s (initial velocity)
£c = 17° (angle)
h = 2.5 ft (initial height)

The parametric equations for projectile motion are:
x = (v0 * cos£c) * t
y = h + (v0 * sin£c) * t - (16 * t^2)

Substituting the values:
x = (136 * cos17°) * t
y = 2.5 + (136 * sin17°) * t - (16 * t^2)

b. To find the time when the ball hits the ground, we need to find the value of t for which y = 0.

Setting y = 0:
0 = 2.5 + (136 * sin17°) * t - (16 * t^2)

Rearranging the equation:
16t^2 - (136 * sin17°) * t - 2.5 = 0

We can solve this quadratic equation by factoring or using the quadratic formula. Let's use the quadratic formula:

The quadratic formula is:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)

For our equation:
a = 16, b = -(136 * sin17°), c = -2.5

Using the quadratic formula, we substitute these values to find t. The positive root will give us the time when the ball hits the ground.

t = [-(136 * sin17°) ± sqrt((136 * sin17°)^2 - 4 * 16 * -2.5)] / (2 * 16)

Now, calculate the value of t.

c. To find how far the ball has traveled when it hits the ground, we need to calculate the value of x at the time t when the ball hits the ground.

Using the value of t we found in part b, substitute it into the x equation:

x = (136 * cos17°) * t

Calculate the value of x.

d. To determine whether the ball clears the 10-foot fence positioned 327 feet from where the ball is hit, compare the height of the ball at the position where it crosses 327 feet with the height of the fence.

Using the x-coordinate of the ball when it reaches 327 feet, substitute this value into the y equation to find the height of the ball at that point.

Compare the height of the ball with the fence height (10 feet) to determine if it clears the fence.