Vertical motion of stone $y=v_{0}\sin\theta t-\frac{1}{2}gt^2$ | Horizontal motion of stone $x=v_{0}\cos\theta\, t$ |

At time of impact $t_{i}$

$0=v_{0}\sin\theta t_{i}-\frac{1}{2}gt_{i}^{2}$

$t_{i}=\frac{2v_{0}\sin\theta}{g}$

$100-10t_{i}=v_{0}\cos\theta\, t_{i}$

$(v_{0}\cos\theta+10)\frac{2v_{0}\sin\theta}{g}=100$

$\frac{2v_{0}^{2}\sin\theta\cos\theta}{g}+\frac{20v_{0}\sin\theta}{g}=100$

$v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}=100g$

$v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}+50=100g+50$

$(v_{0}+\frac{10}{\sqrt{2}})^2=100g+50$

$v_{0}=\frac{-10}{\sqrt{2}}\pm \sqrt{100g+50}$

$v_{0}=25.0 \mathrm{ms^{-1}}$ or $v_{0}=-39.2\mathrm{ms^{-1}}$ but we discard the negative solution.

A. $\tan\theta=\frac{400}{400}$ → $\theta=45^{o}$

B. $v_{BWx}^2+v_{BWy}^2=25$

$v_{BWx}+1=v_{BWy}$

This can either be solved by inspection (ie, if you see that its just a 3,4,5 triangle) or

$2v_{BWx}^2+2v_{BWx}+1=25$

$v_{BWx}^2+v_{BWx}-12=0$

$(v_{BWx}-3)(V_{BWx}+4)=0$

$v_{Bwx}=3$ (negative solution doesn't work) and it follows that $v_{Bwy}=4$

so $\vec{v}_{BW}=(3\hat{i}+4\hat{j})\mathrm{ms^{-1}}$

C. $\tan\phi=\frac{4}{3}=53.13^{o}$

A weight on a scale in a elevator type problem. If the scale reads a smaller number than the actual weight it means that the net force on it is less because it is exerting an upward force on the object in response to additional force exerted on it by the accelerating person. Express the sum of the forces on the scale as $m\vec{g}-m\vec{a}$ to form the equation you can solve for $\vec{a}$.

An inclined plane problem. Find the angle $\theta$ using the equation for the force perpendicular to the plane. You know $m$, you know the maximum acceptable force, and of course you know $g$, so all you need to do is solve for $\theta$.

Draw a diagram in which you sum all the forces on the speaker together. This should make a closed triangle because the net force is zero as the speaker is not accelerating (Newton's Second Law). This will not be a right angled triangle, but will be made up of two adjacent right angled triangles with the angle $\theta$ opposite to $\frac{mg}{2}$ and adjacent to T. the angle $\theta$ can be deduced from $h$ and $l$, and the weight of the object $mg$ is known.

A modified Atwood's Machine..with a twist. You can replace the second weight with the force the window washer exerts. However, if the window washer exerts a force on the rope, what does Newton's Third Law say about the force exerted back on her?

A different modification of Atwood's Machine. (Physicist's are abnormally interested in Atwood's machine!). Replace $m_{2}\vec{g}$ by $m_{A}\vec{g}\sin\theta$. You may note that in the limit $\theta=90^{o}$ the problem reverts to a plain vanilla Atwood's machine.

Did I mention we like Atwood's Machine? Here you can consider various specials cases of the equations we looked at in the lecture. These give you a better intuitive understanding of the system.

Oh, there's a pulley? I guess that means it's Atwood's machine **again**…

Well, actually, no! Recall when we looked at a pulley that was free to move that the force on the object would actually be twice the tension in the rope. So the equation for the net force on $m_{2}$ is going to be a little different. It's gravitational force is balanced by the normal force, so in this case you can say that $m_{2}\vec{a}=2\vec{T}$.

From here on, it *is* like Atwood's machine. Write the equation for the net force on $m_{1}$, which is of course going to be equal to $m_{1}\vec{a}$. Link the motion of the two objects via their common acceleration and the common tension in the string.