A stone is projected at a cliff of height h with an initial speed of 48.0 m/s directed 60.0° above the horizontal. The stone strikes at A, 5.20 s after launching. Find the height, h, of the cliff. Find the speed of the stone just before impact at A. Find the maximum height H reached above the ground.
Height of the cliff, h:
h = 48.0 m/s * 5.20 s * sin(60.0°) = 144 m
Speed of the stone just before impact at A:
v = 48.0 m/s * cos(60.0°) = 24.0 m/s
Maximum height H reached above the ground:
H = 48.0 m/s * 5.20 s * sin(60.0°) - (1/2) * 9.8 m/s^2 * (5.20 s)^2 = 144 m - 131.44 m = 12.56 m
To solve this problem, we can break it down into three parts: finding the height of the cliff (h), the speed of the stone just before impact at A, and the maximum height reached above the ground (H). Let's solve each part step by step.
1. Finding the height of the cliff (h):
The motion of the stone can be divided into horizontal and vertical components. We'll first focus on the vertical component.
The equation for vertical displacement is given by:
h = v0y * t - (1/2) * g * t^2
where:
v0y is the initial vertical velocity component,
t is the time of flight, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
First, we need to find the initial vertical velocity component:
v0y = v0 * sin(theta)
where:
v0 is the initial speed (48.0 m/s), and
theta is the launch angle (60.0°).
v0y = 48.0 m/s * sin(60.0°)
v0y = 48.0 m/s * 0.866
v0y ≈ 41.568 m/s
Next, we can substitute the values into the equation:
h = 41.568 m/s * 5.20 s - (1/2) * 9.8 m/s^2 * (5.20 s)^2
h = (215.5296 - 135.2) m
h ≈ 80.3296 m
Therefore, the height of the cliff (h) is approximately 80.3296 m.
2. Finding the speed of the stone just before impact at A:
The horizontal component of the velocity remains constant throughout the motion. Therefore, the initial horizontal velocity (v0x) will be the final horizontal velocity just before impact.
The equation for horizontal displacement is given by:
x = v0x * t
where:
v0x is the initial horizontal velocity component, and
t is the time of flight.
To find v0x, we can use the equation:
v0x = v0 * cos(theta)
where:
v0 is the initial speed (48.0 m/s), and
theta is the launch angle (60.0°).
v0x = 48.0 m/s * cos(60.0°)
v0x = 48.0 m/s * 0.5
v0x = 24.0 m/s
Since horizontal velocity remains constant, the speed of the stone just before impact at A is equal to the horizontal velocity, hence 24.0 m/s.
3. Finding the maximum height reached above the ground (H):
To find the maximum height, we need to determine the vertical distance travelled by the stone during its flight time.
Using the equation for vertical displacement:
0 = v0y * t - (1/2) * g * t^2
Since we know the time of flight (5.20 s), we can rearrange the equation to solve for the maximum height (H).
(1/2) * g * t^2 = v0y * t
(1/2) * 9.8 m/s^2 * (5.20 s)^2 = 41.568 m/s * 5.20 s
(1/2) * 9.8 m/s^2 * 27.04 s^2 = 215.568 m
Therefore, the maximum height (H) reached above the ground is approximately 215.568 m.
To find the height of the cliff, h, we can apply the equations of motion. The vertical component of the initial velocity can be calculated using the given initial speed and launch angle.
Vertical component of initial velocity (vy) = initial speed * sin( launch angle)
vy = 48.0 m/s * sin(60°)
vy = 48.0 m/s * 0.866
vy = 41.568 m/s
Using the equation of motion for vertical motion, we can find the height of the cliff.
h = vy * time - (1/2) * g * time^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and time is the time taken for the stone to hit the cliff (5.20 seconds).
h = 41.568 m/s * 5.20 s - (1/2) * 9.8 m/s^2 * (5.20 s)^2
h = 215.2576 m - (1/2) * 9.8 m/s^2 * 27.04 s^2
h = 215.2576 m - 133.1776 m
h = 82.08 m
Therefore, the height of the cliff is approximately 82.08 meters.
To find the speed of the stone just before impact at point A, we can use the horizontal component of the initial velocity.
Horizontal component of initial velocity (vx) = initial speed * cos( launch angle)
vx = 48.0 m/s * cos(60°)
vx = 48.0 m/s * 0.5
vx = 24.0 m/s
Since there is no horizontal acceleration (assuming no air resistance), the horizontal component of velocity remains constant. Therefore, the speed of the stone just before impact at point A is 24.0 m/s.
To find the maximum height, H, reached above the ground, we can use the equation of motion for vertical motion to find the time taken to reach maximum height.
Vy = 0 m/s (At maximum height, vertical velocity becomes 0)
Using the equation:
0 = vy - g * time
time = vy / g
time = 41.568 m/s / 9.8 m/s^2
time = 4.24 s
Now we can find the maximum height by substituting this time value into the height equation:
H = vy * time - (1/2) * g * time^2
H = 41.568 m/s * 4.24 s - (1/2) * 9.8 m/s^2 * (4.24 s)^2
H = 176.0112 m - (1/2) * 9.8 m/s^2 * 18.0176 s^2
H = 176.0112 m - 88.0056 m
H = 88.0056 m
Therefore, the maximum height reached above the ground is approximately 88.01 meters.