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MisterNi
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Ok, I edited the post to show all my work, so here it is...
A hole is punched at a height h in the side of a container of height H. The container is full of water. If the water is to shoot as far as possible horizontally, (a) how far from the bottom of the container should the hold be punched? (b) Neglecting frictional losses, how far (initially) from the side of the container will the water land?
Here's what I have so far:
Using Bernoulli's Equation: P + .5*ρ*V^2 + ρ*g*h = a constant.
and since flow rate(Q) = A*V = constant, I set the equations at the opening of the tank = to the hole that's been punched which gives:
P1 + .5*ρ*V1^2 + ρ*g*h1 = P2 + V2^2 + ρ*g*h2
P1 and P2 are 0 since the pressures are atmospheric. .5*ρ*V1^2 is assumed as 0, since the cross sectional area of the tank is much greater than the cross sectional area of the punched hole. I took the initial point at the waterline on top of the tank so ρ*g*h1 is 0 which leaves me with:
.5*ρ*V2^2 = -ρ*g*h2 which simplifies to:
V2 = √[2*g*(H - h)] which is the velocity of the water just as it exists through the punched hole in the tank.
Using the kinematic equation:
d = V2*t + .5*a*t^2
t = √[(2*h)/g]
Here's where I'm stuck. I've tried V2 * t = d and using first order derivatives to determine the maximum...but I get exactly twice the answer in the back of the book which is (H/2). I've posted this problem on another board and a member suggested using the quadratic formula but I couldn't really follow the person's reasoning so I came to the Physics Forums for different suggestions
I'm sure (b) will be easy to find once I can figure out a way to solve for (a). Thanks.
A hole is punched at a height h in the side of a container of height H. The container is full of water. If the water is to shoot as far as possible horizontally, (a) how far from the bottom of the container should the hold be punched? (b) Neglecting frictional losses, how far (initially) from the side of the container will the water land?
Here's what I have so far:
Using Bernoulli's Equation: P + .5*ρ*V^2 + ρ*g*h = a constant.
and since flow rate(Q) = A*V = constant, I set the equations at the opening of the tank = to the hole that's been punched which gives:
P1 + .5*ρ*V1^2 + ρ*g*h1 = P2 + V2^2 + ρ*g*h2
P1 and P2 are 0 since the pressures are atmospheric. .5*ρ*V1^2 is assumed as 0, since the cross sectional area of the tank is much greater than the cross sectional area of the punched hole. I took the initial point at the waterline on top of the tank so ρ*g*h1 is 0 which leaves me with:
.5*ρ*V2^2 = -ρ*g*h2 which simplifies to:
V2 = √[2*g*(H - h)] which is the velocity of the water just as it exists through the punched hole in the tank.
Using the kinematic equation:
d = V2*t + .5*a*t^2
t = √[(2*h)/g]
Here's where I'm stuck. I've tried V2 * t = d and using first order derivatives to determine the maximum...but I get exactly twice the answer in the back of the book which is (H/2). I've posted this problem on another board and a member suggested using the quadratic formula but I couldn't really follow the person's reasoning so I came to the Physics Forums for different suggestions
I'm sure (b) will be easy to find once I can figure out a way to solve for (a). Thanks.
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