A boy kicks a rock off a cliff with a speed of 19.4 m/s at an angle of 52.5° above the horizontal. The rock hits the ground 5.41 s after it was kicked.

a) How high is the cliff? (I found the height to be 60.3 m)

b) What is the speed of the rock right before it hits the ground?

c) What is the maximum height of the rock measured from the top of the cliff?

BUMP!

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To answer these questions, we will need to use equations of motion and the principles of projectile motion. Let's go step-by-step to find the solutions.

a) To find the height of the cliff, we can use the formula for vertical displacement in projectile motion. The equation is:

Δy = v₀y * t - (1/2) * g * t²

where Δy is the vertical displacement, v₀y is the initial vertical component of the velocity, t is the total time of flight, and g is the acceleration due to gravity.

We are given that the rock hits the ground 5.41 s after it was kicked. Therefore, the total time of flight is 5.41 s.

First, we need to find the initial vertical velocity component (v₀y) using the initial velocity (v₀) and the launch angle (θ). We can calculate v₀y using the formula:

v₀y = v₀ * sin(θ)

Given that the initial velocity (v₀) is 19.4 m/s and the launch angle (θ) is 52.5°, we can calculate v₀y as:

v₀y = 19.4 m/s * sin(52.5°) = 14.99 m/s

Now we can substitute the values into the equation for vertical displacement to find the height of the cliff:

Δy = (14.99 m/s) * (5.41 s) - (1/2) * (9.8 m/s²) * (5.41 s)²

Solving this equation, we get:

Δy = 60.3 m

Therefore, the height of the cliff is 60.3 meters.

b) To find the speed of the rock right before it hits the ground, we need to find the final velocity (vf) using the equation:

vf = v₀ + g * t

Given that the initial velocity (v₀) is 19.4 m/s, the acceleration due to gravity (g) is -9.8 m/s² (taking downward as negative), and the total time of flight (t) is 5.41 s, we can calculate vf as:

vf = 19.4 m/s + (-9.8 m/s²) * (5.41 s) = -34.31 m/s (taking downward as negative)

The magnitude of the velocity (speed) is always positive, so the speed of the rock right before it hits the ground is 34.31 m/s.

c) To find the maximum height of the rock measured from the top of the cliff, we can use the equation:

Δy = v₀y² / (2 * g)

Given that the initial vertical velocity component (v₀y) is 14.99 m/s and the acceleration due to gravity (g) is -9.8 m/s² (taking upward as positive), we can calculate the maximum height (Δy) as:

Δy = (14.99 m/s)² / (2 * (-9.8 m/s²)) = 11.5 m

Therefore, the maximum height of the rock measured from the top of the cliff is 11.5 meters.

(b) first

V2x = Vox = ____ m/s i
Vox = 19.4cos52.5

second
Y2 = height of cliff (negative)

You want to find V2y
V2y^2 = Voy^2 + 2ay[Y2-Yo]