You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 22.2 m/s at an angle of 56.0° above the horizontal. If it hits the top of the wall at a speed of 13.2 m/s, how high is the wall?
To determine the height of the wall, we first need to break down the initial velocity of the thrown hook into its horizontal and vertical components.
Given:
Initial velocity (v0) = 22.2 m/s
Launch angle (θ) = 56.0°
Final velocity (vf) = 13.2 m/s
Horizontal component of velocity (v0x):
v0x = v0 * cos(θ)
Vertical component of velocity (v0y):
v0y = v0 * sin(θ)
Now, let's calculate the components of the initial velocity:
v0x = 22.2 m/s * cos(56.0°)
v0y = 22.2 m/s * sin(56.0°)
v0x ≈ 11.21 m/s
v0y ≈ 18.32 m/s
Next, we can use the vertical motion equation to find the height of the wall. We know the final velocity (vf), the initial velocity in the y-direction (v0y), and the acceleration due to gravity (g = 9.8 m/s²).
The vertical motion equation is given as:
vf² = v0y² + 2 * g * h
where h is the height of the wall.
Rearranging the equation, we have:
h = (vf² - v0y²) / (2 * g)
Now, we can substitute the given values into the equation to calculate the height of the wall:
h = (13.2 m/s)² - (18.32 m/s)² / (2 * 9.8 m/s²)
Calculating this, we get:
h ≈ 4.68 m
Therefore, the height of the wall is approximately 4.68 meters.