Questions: D1. During a midterm exam, you toss your pencil straight up in the air and catch it 2.00 s later at the same height as the point of release. (a) What is the acceleration (magnitude and direction) of the pencil at maximum height? (1 mark) pencil is in free-fall a=g downward (b) What is the speed of the pencil at maximum height? (1 mark) Since the pencil is thrown straight up, v=0 at maximum height. (c) Calculate the initial speed of the pencil. (2 marks) ttot=2.00 s ; Δy=0 (returns to initial height) Δy=v0y t+1/2 ay t^2 0=v0y t+1/2 ay t^2 => 0=v0y+1/2 ay t => v0y=-1/2 ay t v0y=-1/2(-g) t v0y=1/2 g t (d) Calculate the maximum height that the pencil reaches. If you did not obtain an answer for (c), use a value of 9.50 m/s. ( 3 marks) vy^2=v0y^2+2 ay Δy vy=0 at max. height 4.90 m Therefore 0=v0y^2+2 ay Δymax => Δymax=-v0y^2/2 ay=-(v0y^2)/2(-g) Δymax=-(9.80 m/s)^2/(2(-9.80 m/s^2))=4.90 m (e) If the pencil was only tossed half as high, how long would it be in the air? If you did not obtain an answer for (d), use a value of 5.00 m . ( 3 marks) Δymax=1/2(4.90 m)=2.45 m vy^2=v0y^2+2 ay Δy 0=v0y^2+2 ay Δymax => v0y=sqrt(-2 ay Δymax)=sqrt(-2(-g)(Δymax)) v0y=sqrt(2 g Δymax)=sqrt(2(9.80 m/s^2)(2.45 m))=6.93 m/s From symmetry of motion, for whole flight vy=-v0y Therefore vy=v0y+ay t => -v0y=v0y-g t t=-2 v0y/g=2(6.93 m/s)/(9.80 m/s^2)= Δytot=v0y tb+t+1/2 ay ttot^2 0=vby ttot+1/2(-g) ttot^2 ttot=-v0y(2)/-g=1.415

Transcript text: D1. During a midterm exam, you toss your pencil straight up in the air and catch it 2.00 s later at the same height as the point of release. (a) What is the acceleration (magnitude and direction) of the pencil at maximum height? (1 mark) pencil is in free-fall \[ \vec{a}=g \text { downward } \] (b) What is the speed of the pencil at maximum height? (1 mark) Since the pencil is thrown straight up, $\vec{v}=0$ at maximum height. O (c) Calculate the initial speed of the pencil. (2 marks) \[ \begin{array}{r} t_{\text {tot }}=2.00 \mathrm{~s} ; \Delta y=0 \quad \text { (returns to initial } \\ \text { height) } \\ \Delta y=v_{0 y} t+\frac{1}{2} a_{y} t^{2} \\ 0=v_{0 y} t+\frac{1}{2} a_{y} t^{2} \Rightarrow 0=v_{0 y}+\frac{1}{2} a_{y} t \Rightarrow v_{0 y}=-\frac{1}{2} a_{y} t \\ v_{0 y}=-\frac{1}{2}(-g) t \\ v_{0 y}=\frac{1}{2} \mathrm{~g} t \end{array} \] (d) Calculate the maximum height that the pencil reaches. If you did not obtain an answer for (c), use a value of $9.50 \mathrm{~m} / \mathrm{s}$. ( 3 marks) \[ v_{y}^{2}=v_{0 y}^{2}+2 a_{y} \Delta y \] $v_{y}=0$ at max. height \[ 4.90 \mathrm{~m} \] \[ \begin{array}{c} \therefore 0=v_{0 y}^{2}+2 a_{y} \Delta y_{\max } \Rightarrow \Delta y_{\max }=-\frac{v_{0 y}^{2}}{2 a_{y}}=\frac{-v_{0 y}^{2}}{2(-\mathrm{g})} \\ \Delta y_{\max }=\frac{-(9.80 \mathrm{~m} / \mathrm{s})^{2}}{2\left(-9.80 \mathrm{~m} / \mathrm{s}^{2}\right)}=4.90 \mathrm{~m} \end{array} \] (e) If the pencil was only tossed half as high, how long would it be in the air? If you did not obtain an answer for (d), use a value of 5.00 m . ( 3 marks) \[ \begin{array}{l} \Delta y_{\max }=\frac{1}{2}(4.90 \mathrm{~m})=2.45 \mathrm{~m} \\ v_{y}^{2}=v_{0 y}^{2}+2 a_{y} \Delta y \\ 0=v_{0 y}^{2}+2 a_{y} \Delta y_{\max } \Rightarrow v_{0 y}=\sqrt{-2 a_{y} \Delta y_{\max }}=\sqrt{-2(-g)\left(\Delta y_{\max }\right)} \\ v_{0 y}=\sqrt{2 g \Delta y_{\max }}=\sqrt{2\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(2.45 \mathrm{~m})}=6.93 \mathrm{~m} / \mathrm{s} \end{array} \] From symmetry of motion, for whole flight $v_{y}=-v_{0 y}$ \[ \begin{aligned} \therefore v_{y} & =v_{0 y}+a_{y} t \Rightarrow-v_{0 y}=v_{0 y}-g t \\ t & =-\frac{2 v_{0 y}}{g}=\frac{2(6.93 \mathrm{~m} / \mathrm{s})}{9.80 \mathrm{~m} / \mathrm{s}^{2}}= \end{aligned} \] \[ \begin{array}{l} \Delta y_{\text {tot }}=v_{0 y} t_{b+t}+\frac{1}{2} a_{y} t_{t o t}^{2} \\ 0=v_{\text {by }} t_{t o t}+\frac{1}{2}(-g) t_{\text {tot }}^{2} \\ t_{\text {tot }}=\frac{-v_{0 y}(2)}{-g}=1.415 \end{array} \] Page 5 of 8