7. A plastic cup, 15.0 cm in height, is filled to the rim with water. A small hole is punctured in the cup at a height 9.00 cm above the bottom of the cup and is allowing water to spew out. At what distance does the water land from the bottom of the cup?(Answer is 14.6 cm)
Use bernoulli's equation to find the velocity of the water leaving when the depth is the full cup. From height of the hole (assume vertical free fall), and velocity, calculate horizontal distance.
h=0.15 m, hₒ=0.09 m.
Bernoulli’s equation between the top surface and the exitting stream:
Pₒ+0+ρghₒ=Pₒ+ρv²/2+ ρgh,
v² = 2g(hₒ-h),
v=sqrt{2g(hₒ-h)}=sqrt{2•9.8•(0.15-0.09)} =1.08m/s.
y=h=gt²/2.
t=sqrt(2h/g) =sqrt(2•0.09/9.8)=0.135 s.
x=vt=1.08•0.135=0.146 m = 14.6 cm
To find the distance at which the water lands from the bottom of the cup, we can use the principle of conservation of energy.
First, let's consider the initial potential energy of the water when it is filled to the rim. The potential energy is given by the formula:
PE_initial = m * g * h
where m is the mass of water, g is the acceleration due to gravity, and h is the height of the cup.
We can calculate the mass of water by using the density of water, which is approximately 1 g/cm³. Given that the cup is filled to the rim, the volume of water is equal to the volume of the cup:
Volume of water = π * r² * h
where r is the radius of the cup. Since the cup is cylindrical, the radius is half the diameter, which is 15.0 cm. Thus, the radius is 7.5 cm.
Substituting the values:
Volume of water = π * (7.5 cm)² * 15.0 cm = 1050π cm³
The mass of water is equal to the volume multiplied by the density:
m = (1050π cm³) * (1 g/cm³) = 1050π g
Now, let's substitute the values into the formula for initial potential energy:
PE_initial = (1050π g) * (9.8 m/s²) * (15.0 cm) = 15075π g cm²/s²
Next, let's consider the final kinetic energy of the water when it lands. At the point of landing, all the potential energy is converted into kinetic energy:
PE_final = KE_final
The kinetic energy is given by the formula:
KE_final = (1/2) * m * v²
where v is the final velocity of the water.
Since the water is spewing out of a small hole, we can assume that its vertical velocity component is zero at the instant of landing (assuming negligible air resistance). This means that the water has only horizontal velocity.
To find the horizontal velocity, we can use kinematic equations. We know that the water has fallen a distance of 9.00 cm before leaving the cup, and the only force acting on it is gravity. The vertical distance the water falls is equal to the change in gravitational potential energy. Using the formula for gravitational potential energy:
PE_initial - PE_final = m * g * Δh
where Δh is the vertical distance, we can rearrange the equation to solve for Δh:
Δh = (PE_initial - PE_final) / (m * g)
Substituting the values:
Δh = (15075π g cm²/s²) / (1050π g * 9.8 m/s²) = 1.5 cm
Since the water is spewing out horizontally, the horizontal distance traveled is the same as the horizontal velocity multiplied by the time it takes to reach the ground. The time it takes to reach the ground can be found using the equation:
Δh = (1/2) * g * t²
Rearranging the equation to solve for time:
t = sqrt((2 * Δh) / g)
Substituting the values:
t = sqrt((2 * 1.5 cm) / 9.8 m/s²) = 0.432 s
Now, let's find the horizontal distance traveled:
Distance = v * t
Since the vertical velocity component is zero, we can use the formula:
Distance = v * t = (horizontal velocity) * t
We don't know the initial horizontal velocity, but we do know that it's the same throughout. So we can write:
Distance = (1/2) * (horizontal velocity) * t
We can find the horizontal velocity from the initial kinetic energy being equal to the final kinetic energy at landing:
(1/2) * m * v² = PE_final
Substituting the values:
(1/2) * (1050π g) * v² = 15075π g cm²/s²
To find the horizontal velocity, v, we can cancel out the g and solve the equation:
(1/2) * (1050π) * v² = 15075π
v² = (15075π) / (1050π) = 14.36
v ≈ sqrt(14.36) = 3.79 cm/s
Finally, let's find the horizontal distance traveled:
Distance = (1/2) * (horizontal velocity) * t = (1/2) * (3.79 cm/s) * (0.432 s) = 0.824 cm
Therefore, the water lands at a distance of approximately 0.824 cm from the bottom of the cup.
To find the distance at which the water lands from the bottom of the cup, we need to consider the initial conditions and analyze the motion of the water as it leaves the cup.
Let's denote the initial height of the water in the cup as "h" and the distance at which the water lands from the bottom as "d."
Given:
- Initial height of water in the cup (h) = 15.0 cm
- Height of the hole from the bottom of the cup (h_hole) = 9.00 cm
To analyze the motion of the water, we can assume that the water stream from the hole behaves as a projectile. This means we can use the equations of motion to solve for the distance d.
The key factor in this problem is that the water is ejected horizontally from the hole, meaning there is no vertical component to its velocity. Thus, the water will travel in a horizontal line from the hole until it lands.
We can use the equation of motion for horizontal motion:
d = v * t
In this case, we need to find the time it takes for the water to land after leaving the hole. To do this, we can use the equation for vertical motion:
h = (1/2) * g * t^2
Since the cup is filled to the rim with water, the initial vertical height (h) is equal to the distance of the hole from the bottom (h_hole).
h = h_hole = 9.00 cm
The acceleration due to gravity (g) is a constant, approximately 9.8 m/s^2, but to ensure consistent units in our calculation, we convert it to cm/s^2 by dividing by 100:
g = 9.8 m/s^2 * (1 m / 100 cm) = 0.098 cm/s^2
Now, we can solve the equation for vertical motion for time (t):
9.00 cm = (1/2) * 0.098 cm/s^2 * t^2
18.00 cm = 0.098 cm/s^2 * t^2
t^2 = (18.00 cm) / 0.098 cm/s^2
t^2 ≈ 183.67 s^2 (approx.)
Taking the square root of both sides:
t ≈ √183.67 s ≈ 13.55 s (approx.)
Now, we can substitute the time (t) into the equation for horizontal motion to find the distance (d):
d = v * t
Since the water is ejected horizontally from the hole, the horizontal velocity component (v) is constant. We can assume the velocity of the water stream is equal to the velocity at which the water leaves the hole.
However, we are not provided with the information about the velocity at which the water leaves the hole. Therefore, we cannot directly calculate the distance (d) without this information.