Answer:
Explanation:
To find the speed of the water as it left the hose, we can use the concept of free fall and the time it takes for the water to hit the ground after the hose is shut off.
Here are the steps to solve the problem:
1. Determine the time it takes for the water to hit the ground after the hose is shut off. In this case, it is given as 2.0 seconds.
2. Use the equation of motion for free fall, which is h = (1/2)gt^2, where h is the vertical displacement, g is the acceleration due to gravity, and t is the time.
3. In this case, the vertical displacement is the height of the hose above the ground, which is 1.5 meters.
4. Rearrange the equation to solve for g, the acceleration due to gravity: g = 2h/t^2.
5. Substitute the values into the equation: g = 2 * 1.5 / (2.0^2) = 1.5 / 4 = 0.375 m/s^2.
6. The speed of the water as it left the hose is equal to the final velocity when the water hit the ground. The final velocity can be found using the equation v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
7. In this case, the initial velocity is 0 m/s because the water starts from rest.
8. Substitute the values into the equation: v = 0 + 0.375 * 2.0 = 0.75 m/s.
Therefore, the speed of the water as it left the hose was 0.75 m/s.
Answer:
[tex] \sf{9.81 \, \sf{m/s}} [/tex]
Explanation:
We can use the kinematic equations for projectile motion to solve this problem. When the water is shut off, the highest point the water reaches will be when its velocity is 0 m/s. From that point, it will fall back to the ground under the influence of gravity.
Let's denote:
- [tex] \sf{v_0} [/tex] as the initial velocity of the water as it leaves the hose,
- [tex] \sf{g} [/tex] as the acceleration due to gravity (approximately [tex] \sf{9.81 \, \sf{m/s}^2} [/tex]),
- [tex] \sf{t} [/tex] as the time it takes for the water to fall from its highest point to the ground (2.0 s),
- [tex] \sf{h} [/tex] as the height above the ground where the hose is pointed (1.5 m).
The water will take the same amount of time to go up to the highest point as it takes to come down from that point. Therefore, the total time for the water to reach the highest point and then hit the ground is [tex] \sf{2t} [/tex].
Using the equation for the displacement of an object under constant acceleration:
[tex] \sf{h = v_0 t - \dfrac{1}{2} g t^2} [/tex]
Since the water starts at a height of 1.5 m, the total displacement until it hits the ground is [tex] \sf{h + 1.5} [/tex]m. We can set the displacement to 0 m at the highest point, so the equation becomes:
[tex] \sf{0 = v_0 t - \dfrac{1}{2} g t^2} [/tex]
Rearranging the equation to solve for [tex] \sf{v_0} [/tex], we get:
[tex] \sf{v_0 = \dfrac{1}{2} g t} [/tex]
Substituting [tex] \sf{g = 9.81 \, \sf{m/s}^2} [/tex] and [tex] \sf{t = 2.0 \, \sf{s}} [/tex], we find:
[tex] \sf{v_0 = \dfrac{1}{2} \times 9.81 \, \sf{m/s}^2 \times 2.0 \, \sf{s}} [/tex]
[tex] \boxed{\bf{v_0 = 9.81 \, \sf{m/s}}} [/tex]
So, the speed of the water as it left the hose was approximately [tex] \bf{9.81 \, \sf{m/s}} [/tex]. Please note that this calculation assumes there is no air resistance and that the water flow is consistent.
[tex] \rule{180pt}{3pt} [/tex]
