# 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 of 32.0 m above sea level, directed at an angle θ above the horizontal with an unknown speed v0.

The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 161.0 m. Assuming that air friction can be neglected, calculate the value of the angle θ.

Calculate the speed at which the rock is launched.

## h = Hi + Vi t - 4.9 t^2

0 = 32 + Vi (6) - 4.9(36)

6 Vi = 144.4

so

Vi = 24.1 m/s up

u = 161/6 = 26.8 horizontal

tan theta = Vi/u

Vo = sqrt (u^2 + Vi^2)

## To find the value of the angle θ, we can use the horizontal distance and the time of flight of the projectile.

Step 1: Find the horizontal component of the initial velocity.

The horizontal component of the initial velocity can be found using the equation:

D = Vx * t, where D is the horizontal distance and t is the time of flight.

Vx = D / t

= 161.0 m / 6.00 s

= 26.8 m/s

Step 2: Find the vertical component of the initial velocity.

The vertical component of the initial velocity can be found using the equation:

H = Vyi * t + 0.5 * g * t^2, where H is the vertical distance, Vyi is the initial velocity in the vertical direction, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time of flight.

Plugging in the values we know:

32.0 m = Vyi * 6.00 s + 0.5 * 9.8 m/s^2 * (6.00 s)^2

We can rearrange the equation to solve for Vyi:

Vyi * 6.00 s = 32.0 m - 0.5 * 9.8 m/s^2 * (6.00 s)^2

Vyi * 6.00 s = 32.0 m - 0.5 * 9.8 m/s^2 * 36.0 s^2

Vyi * 6.00 s = 32.0 m - 176.4 m

Vyi * 6.00 s = -144.4 m

Vyi = -144.4 m / 6.00 s

= -24.1 m/s

Note: The negative sign indicates that the initial velocity in the vertical direction is downward.

Step 3: Calculate the initial velocity.

The initial velocity can be calculated using the horizontal and vertical components of the initial velocity:

v0 = sqrt(Vx^2 + Vyi^2)

= sqrt((26.8 m/s)^2 + (-24.1 m/s)^2)

= sqrt(719.84 m^2/s^2 + 580.81 m^2/s^2)

= sqrt(1300.65 m^2/s^2)

= 36.1 m/s

Therefore, the speed at which the rock is launched is approximately 36.1 m/s.

## To solve this problem, we can use the kinematic equations of motion for projectiles. We'll consider the vertical and horizontal motions separately.

First, let's calculate the initial vertical velocity (v0y) of the rock when it leaves the catapult. We know that the time of flight (t) is 6.00 seconds and the vertical displacement (Δy) is 32.0 m.

Using the equation: Δy = v0y * t + (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Rearranging the equation, we get: v0y = (Δy - (1/2) * g * t^2) / t.

Substituting the values, we have: v0y = (32.0 - (1/2) * 9.8 * (6.00)^2) / 6.00.

Calculating, we find: v0y = 21.6 m/s (rounded to one decimal place).

Next, let's calculate the initial horizontal velocity (v0x) of the rock.

We know that the horizontal displacement (D) is 161.0 m and the time of flight (t) is 6.00 seconds.

The horizontal velocity (v0x) remains constant throughout the entire motion.

Using the equation: D = v0x * t, we can solve for v0x.

Rearranging the equation, we get: v0x = D / t.

Substituting the values, we have: v0x = 161.0 / 6.00.

Calculating, we find: v0x = 26.8 m/s (rounded to one decimal place).

Now, let's calculate the speed at which the rock is launched, which is given by the magnitude of the initial velocity (v0) of the rock.

Using the Pythagorean theorem, we can calculate v0:

v0^2 = v0x^2 + v0y^2.

Substituting the values, we have: v0^2 = (26.8)^2 + (21.6)^2.

Calculating, we find: v0 ≈ 34.3 m/s (rounded to two decimal places).

Finally, to calculate the value of the angle θ, we use the equation: tan(θ) = v0y / v0x.

Substituting the values, we have: tan(θ) = 21.6 / 26.8.

Calculating, we find: θ ≈ 38.5 degrees (rounded to one decimal place).

Therefore, the angle θ is approximately 38.5 degrees and the rock is launched at a speed of approximately 34.3 m/s.