In the figure below, a stone is projected at a cliff of height h with an initial speed of 46.0 m/s directed 60.0° above the horizontal. The stone strikes at A, 5.30 s after launching.

(a) Find the height, h, of the cliff.
m

(b) Find the speed of the stone just before impact at A.
m/s

(c) Find the maximum height H reached above the ground.
m

To solve this problem, we need to consider the motion of the stone both horizontally and vertically. Let's break it down step by step.

(a) Finding the height, h, of the cliff:
Using the vertical motion of the stone, we can use the equation of motion: h = v₀y × t + (1/2) × g × t²
Here, v₀y is the vertical component of the initial velocity, t is the time taken to hit the cliff, and g is the acceleration due to gravity (which is approximately 9.8 m/s²).

First, we need to find the vertical component of the initial velocity, v₀y. We can do this by using the formula: v₀y = v₀ × sin(θ)
Given that the initial speed, v₀, is 46.0 m/s and the launch angle, θ, is 60.0°, we can calculate v₀y.

v₀y = 46.0 m/s × sin(60.0°)
v₀y = 39.80 m/s (rounded to two decimal places)

Now, using the formula for height, we can plug in the known values:
h = (39.80 m/s) × (5.30 s) + (1/2) × (9.8 m/s²) × (5.30 s)²
h = 210.18 m (rounded to two decimal places)

Therefore, the height of the cliff, h, is approximately 210.18 meters.

(b) Finding the speed of the stone just before impact at A:
We can find the vertical component of the final velocity, vy, just before impact using the equation: vy = v₀y + g × t.
Given that v₀y is 39.80 m/s and t is 5.30 seconds, we can calculate vy.

vy = 39.80 m/s + (9.8 m/s²) × (5.30 s)
vy = 92.34 m/s (rounded to two decimal places)

Next, we can find the horizontal component of the final velocity, vₓ. Since there is no horizontal acceleration, the horizontal component of velocity remains constant throughout the motion. Therefore, vₓ is equal to the horizontal component of the initial velocity, which is v₀x.

v₀x = v₀ × cos(θ)
Given that v₀ is 46.0 m/s and the launch angle, θ, is 60.0°, we can calculate v₀x.

v₀x = 46.0 m/s × cos(60.0°)
v₀x = 23.00 m/s (rounded to two decimal places)

Finally, we can find the magnitude of the final velocity, v, using the Pythagorean theorem:
v = sqrt(vₓ² + vy²)
v = sqrt((23.00 m/s)² + (92.34 m/s)²)
v = 96.33 m/s (rounded to two decimal places)

Therefore, the speed of the stone just before impact at A is approximately 96.33 m/s.

(c) Finding the maximum height H reached above the ground:
To find the maximum height, H, we need to consider the vertical motion of the stone. At the maximum height, the vertical component of velocity, vy, becomes zero.

Using the equation of motion: vy = v₀y - g × t
Setting vy to zero, we can find the time it takes for the stone to reach the maximum height.

0 = 39.80 m/s - (9.8 m/s²) × t
t = 4.08 s (rounded to two decimal places)

Now, we can use this time in the height equation: H = v₀y × t - (1/2) × g × t²
Let's calculate:

H = (39.80 m/s) × (4.08 s) - (1/2) × (9.8 m/s²) × (4.08 s)²
H = 81.45 m (rounded to two decimal places)

Therefore, the maximum height, H, reached above the ground is approximately 81.45 meters.

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