An arrow is shot from a height of 1.3 m toward a cliff of height H. It is shot with a velocity of 28 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.8 s later.
To find the height of the cliff (H), we can use the equations of motion for projectile motion.
Step 1: Split the initial velocity into horizontal and vertical components.
The horizontal component (Vx) can be found using the equation Vx = V * cos(θ), where V is the magnitude of the initial velocity (28 m/s) and θ is the angle above the horizontal (60°).
Vx = 28 m/s * cos(60°)
Step 2: Find the time it takes for the arrow to reach the top of the cliff.
The time of flight can be found using the equation t = 2 * Vy / g, where Vy is the vertical component of the velocity and g is the acceleration due to gravity (9.8 m/s^2).
Since the arrow lands on the top edge of the cliff, the total time of flight is 3.8 seconds.
3.8 s = 2 * Vy / 9.8 m/s^2
Step 3: Find the maximum height reached by the arrow.
Using the equation y = yo + Voy * t - 0.5 * g * t^2, where y is the vertical distance, yo is the initial displacement (1.3 m), Voy is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Since the arrow starts at a height of 1.3 m, the initial vertical displacement (yo) is 1.3 m.
The initial vertical velocity (Voy) can be found using the equation Voy = V * sin(θ).
Voy = 28 m/s * sin(60°)
Now, let's calculate the values:
Vx = 28 m/s * cos(60°)
= 28 m/s * 0.5
= 14 m/s
3.8 s = 2 * Vy / 9.8 m/s^2
Vy = 3.8 s * 9.8 m/s^2 / 2
= 18.62 m/s
Voy = 28 m/s * sin(60°)
= 28 m/s * 0.866
= 24.248 m/s
Using the equation y = yo + Voy * t - 0.5 * g * t^2, we can find the maximum height reached:
y = 1.3 + 24.248 * 3.8 - 0.5 * 9.8 * (3.8)^2
= 1.3 + 92.6152 - 70.3248
= 23.5904 m
Therefore, the height of the cliff (H) is approximately 23.5904 meters.