Finding Constants Using Continuity Conditions

[glebovg]

Homework Statement

A ball falls from rest at a height H above a lake. Let y = 0 at the surface of the lake. As the ball falls, it experiences a gravitational force -mg. When it enters the water, it experiences a buoyant force B so the net force in the water is B - mg.

a) Write an expression for v(t) and y(t) while the ball is falling in air.
b) In the water, let v₂(t) = at + b and y₂(t) = (1/2)at² + bt + c where a = (B - mg)/m. Use
continuity conditions at the surface of the water to find the constants b and c.

The attempt at a solution

a)
Since $\vec{v}=\int\vec{a}dt=\vec{a}t+\vec{v}_{0}$ and the initial velocity is 0, we have $\vec{v}(t)=\vec{a}t$. Using $\vec{F}=m\vec{a}$ yields $v(t)=-gt$.
Also, $\vec{r}=\int\vec{v}dt=\frac{1}{2}\vec{a}t^{2}+\vec{v}t+\vec{r}_{0}$. Again, since the initial velocity is 0 and $y=r-r_{0}$ we have $y(t)=-\frac{1}{2}gt^{2}$.

 

[SammyS]

Homework Statement

A ball falls from rest at a height H above a lake. Let y = 0 at the surface of the lake. As the ball falls, it experiences a gravitational force -mg. When it enters the water, it experiences a buoyant force B so the net force in the water is B - mg.

a) Write an expression for v(t) and y(t) while the ball is falling in air.
b) In the water, let v₂(t) = at + b and y₂(t) = (1/2)at² + bt + c where a = (B - mg)/m. Use
continuity conditions at the surface of the water to find the constants b and c.

The attempt at a solution

a)
Since $\vec{v}=\int\vec{a}dt=\vec{a}t+\vec{v}_{0}$ and the initial velocity is 0, we have $\vec{v}(t)=\vec{a}t$. Using $\vec{F}=m\vec{a}$ yields $v(t)=-gt$.
Also, $\vec{r}=\int\vec{v}dt=\frac{1}{2}\vec{a}t^{2}+\vec{v}t+\vec{r}_{0}$. Again, since the initial velocity is 0 and $y=r-r_{0}$ we have $y(t)=-\frac{1}{2}gt^{2}$.

Isn't the ball at y = H, at time t=0 ?

 

[glebovg]

Yes. So the last equation should be $y(t)=-\frac{1}{2}gt^{2} + H$. Can you help with b)? I do not understand the question.

 

[SammyS]

Yes. So the last equation should be $y(t)=-\frac{1}{2}gt^{2} + H$. Can you help with b)? I do not understand the question.

At what time does the ball enter the water?

 

[glebovg]

When y = 0, $-\frac{1}{2}gt^{2}+H=0$ and so $t=\sqrt{\frac{2H}{g}}$.

 

[glebovg]

I do not understand the question. Please help.