A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D. The hose shoots the water at 45∘ above the horizontal from the same level as the base of the tank and is a distance 6D away.
For what range of launch speeds (v0) will the water enter the tank? Ignore air resistance, and express your answer in terms of D and g
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To determine the range of launch speeds for the water to enter the tank, we first need to determine the height at which the water enters the tank.
From the problem, we can see that the water is shot at 45 degrees above the horizontal. This means that the initial velocity of the water can be broken down into two components: one along the horizontal direction and one along the vertical direction.
Since the water is shot from the same level as the base of the tank, the vertical component of the initial velocity will determine the height at which the water enters the tank.
The vertical component of the initial velocity can be found using trigonometry:
v_y = v_0 * sin(45)
= (v_0 * √2) / 2
The time it takes for the water to reach the tank can be found using the vertical component of the initial velocity and the acceleration due to gravity: g.
y = v_y * t - (1/2) * g * t^2
= ((v_0 * √2) / 2) * t - (1/2) * g * t^2
The height of the tank is 2D. Therefore, we can set y = 2D and solve for t:
2D = ((v_0 * √2) / 2) * t - (1/2) * g * t^2
Simplifying the equation:
0 = - (1/2) * g * t^2 + ((v_0 * √2) / 2) * t - 2D
(1/2) * g * t^2 - ((v_0 * √2) / 2) * t + 2D = 0
We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = (1/2) * g, b = ((v_0 * √2) / 2), and c = 2D.
Substituting the values:
t = ( -((v_0 * √2) / 2) ± √(((v_0 * √2) / 2)^2 - 4 * (1/2) * g * 2D) ) / (2 * (1/2) * g)
Simplifying further:
t = ( -((v_0 * √2) / 2) ± √((v_0^2 * 2) / 4 - 4gD) ) / g
t = ( -((v_0 * √2) / 2) ± √((v_0^2) / 2 - 4gD) ) / g
Since we want the water to enter the tank, t should be positive. Therefore, we only consider the positive root:
t = ( -((v_0 * √2) / 2) + √((v_0^2) / 2 - 4gD) ) / g
To simplify the equation further, we multiply the numerator and denominator by √2:
t = ( -v_0 + √(v_0^2 - 8gD) ) / (2g√2)
Since the time it takes for the water to reach the tank is t = 6D/v_0, we can replace t in the equation:
6D/v_0 = ( -v_0 + √(v_0^2 - 8gD) ) / (2g√2)
Multiplying both sides by v_0:
6D = (-v_0^2 + v_0 √(v_0^2 - 8gD) ) / (2√2)
Multiplying both sides by (2√2):
12√2D = (-v_0^2 + v_0 √(v_0^2 - 8gD) )
Rearranging the equation:
v_0^2 - v_0 √(v_0^2 - 8gD) - 12√2D = 0
Now, we can solve this quadratic equation for v_0. However, since there are no values for D and g in the problem, we cannot find the exact range of launch speeds without specific values for D and g.