In the figure below, a stone is projected at a cliff of height h with an initial speed of 44m/s directed 60° above the horizontal. The stone strikes at A, 5.8s after launching.
(a) Find the height, h, of the cliff.
(b) Find the speed of the stone just before impact at A.
(c) Find the maximum height H reached above the ground.
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To answer these questions, we can use the equations of motion for projectile motion, which describe the motion of an object in two dimensions under the influence of gravity. The key equations we need are:
1. The horizontal displacement equation: x = v₀⋅cosθ⋅t
2. The vertical displacement equation: y = v₀⋅sinθ⋅t - 1/2⋅g⋅t²
3. The vertical velocity equation: vy = v₀⋅sinθ - g⋅t
4. The time of flight equation: t = 2⋅v₀⋅sinθ / g
Now, let's solve each part of the problem:
(a) To find the height h of the cliff, we need to find the vertical displacement of the stone when it strikes at point A. We are given the time of flight as 5.8s. Substituting the values into the vertical displacement equation, we have:
y = v₀⋅sinθ⋅t - 1/2⋅g⋅t²
= 44⋅sin(60°)⋅5.8 - 1/2⋅9.8⋅(5.8)²
≈ 150.35 - 162.09
≈ -11.74 m
Since the stone strikes at A, the vertical displacement at A is zero. Therefore, we can set y = 0 and solve for h:
0 = 44⋅sin(60°)⋅t - 1/2⋅9.8⋅t²
0 = 44⋅sin(60°)⋅5.8 - 1/2⋅9.8⋅(5.8)²
11.74 = 169.92 - 1/2⋅9.8⋅(5.8)²
1/2⋅9.8⋅(5.8)² = 169.92 - 11.74
1/2⋅9.8⋅(5.8)² = 158.18
h = 158.18 ≈ 158.2 m
Therefore, the height of the cliff is approximately 158.2 meters.
(b) To find the speed of the stone just before impact at A, we can use the vertical velocity equation. At point A, the stone has a vertical velocity of 0, since it reaches the highest point in its trajectory. So, we can set vy = 0 and solve for v₀:
vy = v₀⋅sinθ - g⋅t
0 = 44⋅sin(60°) - 9.8⋅5.8
9.8⋅5.8 = 44⋅sin(60°)
v₀ = (9.8⋅5.8) / sin(60°)
v₀ ≈ 10.92 / 0.866
v₀ ≈ 12.6 m/s
Therefore, the speed of the stone just before impact at A is approximately 12.6 m/s.
(c) To find the maximum height H reached above the ground, we need to find the vertical displacement at the highest point of the stone's trajectory. Using the time of flight equation, we can calculate the time it takes for the stone to reach its highest point:
t = 2⋅v₀⋅sinθ / g
= 2⋅44⋅sin(60°) / 9.8
≈ 100.8 / 9.8
≈ 10.29 s
Substituting this value into the vertical displacement equation, we have:
y = v₀⋅sinθ⋅t - 1/2⋅g⋅t²
= 44⋅sin(60°)⋅10.29 - 1/2⋅9.8⋅(10.29)²
≈ 380.35 - 508.29
≈ -127.94 m
Since the maximum height is the vertical displacement above the ground, we take the absolute value:
H = |y| = 127.94 m
Therefore, the maximum height reached by the stone above the ground is approximately 127.94 meters.