A projectile is fired into the air from the top of a cliff of height h = 186 m above a valley. Its initial velocity is v0 = 64.8 m/s at an angle θ = 59° above the horizontal. Where does the projectile land? (Ignore any effects due to air resistance.)

ho = 186 m.

Vo = 64.8m/s[59o]
Xo = 64.8*Cos59 = 33.37 m/s
Yo = 64.8*sin59 = 55.54 m/s

Tr = -Yo/g = -55.54/-9.8 = 5.67 s. = Rise time.

h = ho + (-Yo^2)/2g
h = 186 + -(55.54^2)/-19.6 = 343.4 m. Above gnd.

h = 0.5g*t^2 = 343.4 m.
4.9t^2 = 343.4
t^2 = 70.1
Tf = 8.37 s. = Fall time.

d = Xo*(Tr+Tf) = 33.37m/s * (5.67+8.37)s
= 469 m. = Hor. distance.

To determine where the projectile lands, we need to analyze the motion of the projectile and calculate its horizontal range.

The horizontal range is the horizontal distance covered by the projectile before it hits the ground. To find the range, we can split the initial velocity into horizontal and vertical components.

First, let's find the initial vertical velocity component (Vy₀) and the initial horizontal velocity component (Vx₀).

Given:
Initial velocity (v₀) = 64.8 m/s
Launch angle (θ) = 59°

To find Vy₀:
Vy₀ = v₀ * sin(θ)

To find Vx₀:
Vx₀ = v₀ * cos(θ)

Substituting the given values into the formulas, we get:

Vy₀ = 64.8 m/s * sin(59°)
Vx₀ = 64.8 m/s * cos(59°)

Now, we can analyze the motion of the projectile in the vertical direction to determine the time it takes to reach the ground.

The equation for displacement in the vertical direction is:

y = y₀ + Vy₀ * t - (1/2) * g * t²

Where:
y₀ = initial vertical position (height of the cliff) = 186 m
Vy₀ = initial vertical velocity
g = acceleration due to gravity = 9.8 m/s²
t = time

When the projectile hits the ground, its vertical displacement (y) will be 0. Therefore, we can rewrite the equation as:

0 = 186 m + Vy₀ * t - (1/2) * g * t²

Simplifying the equation, we get:

(1/2) * g * t² - Vy₀ * t - 186 m = 0

This is a quadratic equation in terms of t. We can solve it using the quadratic formula:

t = (-b ± sqrt(b² - 4ac)) / (2a)

Where:
a = (1/2) * g
b = -Vy₀
c = -186 m

Substituting the values, we get:

t = (-(-Vy₀) ± sqrt((-Vy₀)² - 4 * (1/2) * g * (-186 m))) / (2 * (1/2) * g)

Simplifying further:

t = (Vy₀ ± sqrt(Vy₀² + 2 * g * 186 m)) / g

Knowing the time of flight, we can calculate the horizontal range using the equation:

Range = Vx₀ * t

Substituting the values we obtained earlier:

Range = (64.8 m/s * cos(59°)) * t

Now, we have all the necessary equations to calculate the range. Let's proceed with the calculations.