Water is leaving a hose at 6.8 m/s. If the target is 2 m away horizontally, What angle should the water have initially?
Require that the range,
R = Vo^2*sin(2A)/g = 2.0 m
Solve for the angle A. There will be two possible answers.
Water
To find the angle at which the water should be initially directed, we can use trigonometry. We have two known values: the horizontal distance (2 m) and the initial velocity of the water (6.8 m/s).
Let's assume the initial angle of the water from the hose is θ. We need to find the value of θ.
Here's how we can solve it using trigonometry:
1. In a projectile motion, we can split the initial velocity (6.8 m/s) into its horizontal and vertical components. The horizontal component (Vx) remains constant, while the vertical component (Vy) changes due to gravity.
2. The horizontal component (Vx) of the velocity stays the same throughout the motion. Since the only horizontal force acting on the water is gravity (which is negligible for horizontal motion), the horizontal velocity component remains constant. So, Vx = 6.8 m/s.
3. The vertical component (Vy) of the velocity changes due to gravity. At the maximum height, the vertical velocity becomes zero (Vy = 0). The time taken to reach the maximum height (t) can be found using the vertical component equation:
Vy = Voy - g * t (where Voy is the initial vertical velocity and g is the acceleration due to gravity)
Since the vertical component initially is Vy = Voy (as it starts from the ground), the equation simplifies to:
Voy = g * t
4. The time taken to reach the maximum height is equal to the total time of flight divided by 2. Since the initial vertical velocity is equal to the final vertical velocity at maximum height, the total time of flight is given by:
t_total = 2 * t
(The factor of 2 arises because the time to reach the maximum height is equal to the time taken to descend from the maximum height to the ground)
5. The total time of flight can be calculated using the horizontal component formula:
2 * t = 2 * (Vx / g)
(where g is the acceleration due to gravity, approximately 9.8 m/s^2)
6. Substitute the value of Vx from step 2 into the equation:
2 * t = 2 * (6.8 / 9.8)
7. Simplify the equation:
t = 6.8 / 9.8
8. Now, we can find the initial vertical velocity (Voy) using the equation from step 3:
Voy = g * t
Voy = 9.8 * (6.8 / 9.8)
9. Simplify the equation:
Voy = 6.8 m/s
10. Now, we can use trigonometry to find the angle θ:
tan(θ) = Vy / Vx
tan(θ) = 6.8 / 6.8
tan(θ) = 1
θ = tan^(-1)(1)
θ = 45 degrees
Therefore, the water should be initially directed at an angle of 45 degrees above the horizontal.