Suppose that you hold a garden hose on the ground. It spurts water at a speed of 6,0m.s' (a) at what angle (to the horizontal) should you hold the hose pipe if you want to water plants 2m away? (b) explain why there are two possible angles?.
(V^2/g)*sin2A = 2.0 m
is the equation for where the water lands.
It is the same equation that applies to ballistics (cannonballs, thrown rocks etc)
Solve for the angle A. There will be two of them, since 2A can range from 0 to 180 degrees.
A = 16.5 and 73.5 degrees
Note that they are complementary angles.
To determine the angle at which you should hold the hose pipe, we can use trigonometry. The two angles mentioned in part (b) are due to the fact that there are two possible solutions to the problem.
(a) To find the angle, we need to consider the components of the velocity. The horizontal component of the velocity will determine how far the water travels along the ground, and the vertical component will determine the height at which the water reaches the plants.
Let's assume we want the water to hit the plants 2m away. We can divide this problem into two parts: horizontal motion and vertical motion.
Horizontal motion:
The horizontal distance covered (2m) can be calculated using the horizontal velocity (6.0 m/s) and the time of flight (t). Since the velocity is constant, we can use the formula d = v * t. Rearranging it for time, we get t = d / v.
Using the given values, t = 2m / 6.0m/s = 1/3 s.
Vertical motion:
The vertical distance will depend on the acceleration due to gravity (g), the time of flight (t), and the initial vertical velocity (u), which is 0 as the water starts from rest in the vertical direction. The formula to calculate the vertical distance is d = u * t + (1/2) * g * t^2.
Since u = 0, the formula simplifies to d = (1/2) * g * t^2.
The vertical distance we want to achieve is 0 (as the water should reach the ground) at time t = 1/3s. So, d = 0 = (1/2) * g * (1/3s)^2. Rearranging the equation, we get g = -18 m/s^2 (negative sign indicates downward acceleration due to gravity).
Now, we can determine the angle using trigonometry.
tanθ = (vertical component of velocity)/(horizontal component of velocity) = (g * t)/v
tanθ = ( -18 * (1/3) )/( 6.0 )
tanθ = -0.1
θ = tan^(-1)(-0.1)
θ ≈ -5.71° (It is important to note that this negative angle indicates below the horizontal.)
There is another possible angle as well, tanθ = -( -0.1 ) = 0.1.
θ = tan^(-1)(0.1)
θ ≈ 5.71°
Therefore, the two possible angles are approximately -5.71° and 5.71°.
(b) The reason why there are two possible angles is due to the symmetry of the problem. Water can reach the same height and distance with the hosepipe inclined above or below the horizontal. So, the two angles represent the inclination above and below the horizontal from which the water reaches the desired target.