a fire hose held near the ground shoots water at a speed of 6.4m/s at what angle should the nozzle point in order that the water lands 2.5m away? And why are there two different angles?
To determine the angle at which the nozzle should point, we can use the principles of projectile motion. The water from the fire hose will follow a parabolic trajectory.
First, let's break down the problem into horizontal and vertical components. The horizontal component represents the distance traveled by the water (2.5m), while the vertical component represents the height achieved by the water.
1. Horizontal component:
The horizontal speed of the water is given as 6.4 m/s. We know that the horizontal distance is 2.5m, so we can use the formula: distance = speed × time. Since the speed remains constant, we can rearrange the formula to time = distance / speed. Plugging in the values, we get: time = 2.5m / 6.4m/s = 0.39s.
2. Vertical component:
To find the angle, we need to consider the initial vertical velocity of the water. Since the water is initially shot from the same height as the nozzle (near the ground), the vertical velocity is zero at that point. The only force acting on the water is gravity.
Using the equation for vertical motion, we can calculate the time it takes for the water to reach its maximum height and hit the ground. The formula is: time = 2 × (vertical distance / gravity)^0.5, where gravity is approximately 9.8 m/s^2. In our case, the vertical distance is the same as the nozzle's height. Plugging in the values, we get: time = 2 × (h / 9.8)^0.5.
3. Combining horizontal and vertical components:
Since the total time is the same for both components (vertical and horizontal), we can equate the time in the vertical equation to the time calculated in the horizontal equation.
time = 2 × (h / 9.8)^0.5 = 0.39s.
We can solve this equation to find the height (h) above the ground. Once we have the height, we can calculate the angle using trigonometry.
Now, why are there two different angles?
In projectile motion, there are two possible angles for the same range. These angles are symmetrical, resulting in the same horizontal displacement. One angle corresponds to the case when the water goes up first and then falls to the ground (a more arched trajectory), while the other angle corresponds to the case when the water immediately falls to the ground (a flatter trajectory). Hence, for a given range, there are typically two angles that lead to the same result.
the range equation is ... R = [v^2 * sin(2Θ)] / g
... v is the launch velocity
... Θ is the launch angle
... g is gravitational acceleration
plug in the given values and solve for Θ
there are two solutions
... a higher, looping trajectory
... and a lower, flatter trajectory