A projectile is launched off of a 18.6 m high cliff with a velocity of 12.8 m/s. If the projectile is launched at an angle of 20° above the horizontal, how far did it travel horizontal before it hit the ground below?
To find the horizontal distance traveled by the projectile before hitting the ground, we can break down the initial velocity into horizontal and vertical components.
1. Calculate the horizontal component of the initial velocity:
The horizontal component can be found using the formula:
Vx = V * cos(θ)
where Vx is the horizontal component of the velocity, V is the magnitude of the velocity (12.8 m/s), and θ is the launch angle (20°).
Vx = 12.8 m/s * cos(20°)
2. Calculate the vertical component of the initial velocity:
The vertical component can be found using the formula:
Vy = V * sin(θ)
where Vy is the vertical component of the velocity, V is the magnitude of the velocity (12.8 m/s), and θ is the launch angle (20°).
Vy = 12.8 m/s * sin(20°)
3. Calculate the time of flight:
The time of flight can be found using the formula:
T = 2 * Vy / g
where T is the time of flight, Vy is the vertical component of the velocity, and g is the acceleration due to gravity (approximately 9.8 m/s²).
T = 2 * 12.8 m/s * sin(20°) / 9.8 m/s²
4. Calculate the horizontal distance traveled:
The horizontal distance can be found using the formula:
D = Vx * T
where D is the horizontal distance, Vx is the horizontal component of the velocity, and T is the time of flight.
D = (12.8 m/s * cos(20°)) * (2 * 12.8 m/s * sin(20°) / 9.8 m/s²)
5. Solve for D:
Calculate the value of D using the above equation.
D ≈ 11.59 meters
Therefore, the projectile traveled approximately 11.59 meters horizontally before hitting the ground below.