Water emerges vertically from a hose at ground level and rises to a height of 3.2 m.
(a) At what speed does the water emerge from the nozzle?
(b) How long is a given drop in the air?
(1/2) m Vi^2 = m g h
so
Vi = sqrt(2 g h)
v = Vi - g t
0 = Vi - 9.81 t
solve for t, time to rise to 3.2 m
then time in air = 2 t
How do I solve it in another way and please give it in full
To find the speed at which the water emerges from the nozzle, we can use the principle of conservation of energy. The potential energy of the water at ground level is converted to kinetic energy as it rises to a height of 3.2 m.
(a) Let's use the equation for potential energy,
Potential Energy = Mass x Gravity x Height
Since we are interested in the speed at which the water emerges, we need to find the velocity. The equation for kinetic energy is,
Kinetic Energy = 1/2 x Mass x Velocity^2
Setting the initial potential energy equal to the final kinetic energy, we have:
Mass x Gravity x Height = 1/2 x Mass x Velocity^2
Since the mass of the water cancels out, we have:
Gravity x Height = 1/2 x Velocity^2
Rearranging the equation to solve for velocity, we get:
Velocity = sqrt(2 x Gravity x Height)
To find the speed, plug in the values:
Velocity = sqrt(2 x 9.8 m/s^2 x 3.2 m)
Calculating the expression, we get:
Velocity = sqrt(62.72) ≈ 7.92 m/s
So, the water emerges from the nozzle at a speed of approximately 7.92 m/s.
(b) To find how long a given drop is in the air, we can use the equation of motion for vertical displacement under constant acceleration. The equation is:
Displacement = Initial Velocity x Time + 1/2 x Acceleration x Time^2
In this case, the initial velocity is 0 m/s since the water begins from rest. The acceleration is the acceleration due to gravity, which is -9.8 m/s^2 (negative sign because it acts downwards). We are interested in the displacement, which is the height the water rises (3.2 m).
Using this information, we can rearrange the equation to solve for time:
3.2 m = 0 m/s x Time + 1/2 x (-9.8 m/s^2) x Time^2
Simplifying the equation, we have:
3.2 m = -4.9 m/s^2 x Time^2
Dividing both sides by -4.9 m/s^2, we get:
Time^2 = -3.2 m / -4.9 m/s^2
Calculating the expression, we get:
Time^2 ≈ 0.653 m/s^2
Taking the square root of both sides, we find:
Time ≈ sqrt(0.653) ≈ 0.81 s
So, a given drop in the air remains in the air for approximately 0.81 seconds.