A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 32.0 m above sea level, directed at an angle theta = 46.9° above the horizontal, and with a speed v = 29.6 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.

Vo = 29.6m/s[46.9o].

Xo = 29.6*Cos46.9. = m/s.
Yo = 29.6*sin46.9 = m/s.

Y = Yo + g*Tr.
Y = 0.
g = -9.8 m/s^2.
Tr = Rise time.
Tr = ?

Y^2 = Yo^2 + 2g*h.
Y = 0.
h = ?

0.5g*Tf^2 = h + 32.
g = 9.8 m/s^2.
Tf = Fall time.
Tf = ?

D = Xo*(Tr+Tf).

To find the horizontal distance traveled by the projectile, we need to analyze the motion of the rock both horizontally and vertically.

We can start by breaking down the initial velocity of the rock into its horizontal and vertical components.

The vertical component of the initial velocity can be found using the formula: vy = v * sin(theta)

where vy is the vertical component of velocity, v is the initial velocity, and theta is the launch angle.

Plugging in the given values: v = 29.6 m/s and theta = 46.9°, we have:

vy = 29.6 m/s * sin(46.9°)
= 29.6 m/s * 0.72
= 21.312 m/s

Next, we can find the time it takes for the rock to reach its maximum height. At this point, the vertical component of velocity becomes zero. We can use the equation:

vy = vy0 + (-g * t)

where vy is the final vertical velocity (which is zero at maximum height), vy0 is the initial vertical velocity, g is the acceleration due to gravity (approximately -9.8 m/s^2), and t is the time.

Rearranging the equation to solve for t, we have:

t = (vy - vy0) / (-g)
= (0 - 21.312 m/s) / (-9.8 m/s^2)
= -21.312 m/s / (-9.8 m/s^2)
= 2.174 s

Now, we can use the time it takes to reach maximum height to find the total time of flight. Since the projectile will take the same amount of time to reach its maximum height as it will to fall back down, the total time of flight is twice the time to reach maximum height.

Total time of flight (T) = 2 * t
= 2 * 2.174 s
= 4.348 s

Finally, we can calculate the horizontal distance traveled by the projectile using the formula:

D = vx * T

where D is the horizontal distance, vx is the horizontal component of velocity (which remains constant throughout the projectile's motion), and T is the total time of flight.

The horizontal component of the initial velocity (vx) can be found using the formula: vx = v * cos(theta)

Plugging in the given values: v = 29.6 m/s and theta = 46.9°, we have:

vx = 29.6 m/s * cos(46.9°)
= 29.6 m/s * 0.66
= 19.536 m/s

Now, we can calculate the horizontal distance D:

D = vx * T
= 19.536 m/s * 4.348 s
= 84.87 m

Therefore, the horizontal distance traveled by the projectile is approximately 84.87 meters.