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$\begin{array}{l}{\text { A physic sooksides off a horizontal tabletop with a speed of } 1.1 \mathrm{m} / \mathrm{s} \text { it strikes the floor in } 0.350 \mathrm{s} \text { . Ignore air resistance. Find }} \\ {\text { (a) the height of the tabletop above the floor; }} \\ {\text { (b) I the heizontal distane-from the edge of the table to the point where the book strikes the floor; }} \\ {\text { (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before }} \\ {\text { the book reaches the floor. }} \\ {\text { (d) Draw } x \text { -t, } y-t \text { , V } x \text { -tand } v \text { -t - graphs for the motion. }}\end{array}$

$\text { X- component }$

$V_{0 x}=1.1 \mathrm{m} / \mathrm{s} \longrightarrow a_{x}=0$

$t=0.35 \mathrm{s}$

$y-\text { component }$

$a_ y=-9.8 \mathrm{m} / \mathrm{s}^2$

$V_{0y}=0$

$t=0.35 \mathrm{s}$

$y-y_{0}=v_{0y} t+\frac{1}{2} a_ y t^{2}$

$y-y_{0}=0+\frac{1}{2}(-9.8)(0.35)^{2}$

$=\quad-0.6 \mathrm{m}$

$x-x_{0}=V_{0 x} t+\frac{1}{2} a_{x} t^{2}$

$x-x_{0}=1.1(0.35)+0=0.385 \mathrm{m}$

$V_{x}=V_{0 x}+a_{x} t=1.1 \mathrm{m} / \mathrm{s}$

$V_{y}=V_{0y}+a_{y} t=-3.43 \mathrm{m} / \mathrm{s}$

$V=\sqrt{{v_{x}}^{2}+{v_{y}}^{2}}$

$=\sqrt{(1.1)^{2}+(-3 \cdot 43)^{2}}=3 \cdot 6 \mathrm{m} / \mathrm{s}$

$\alpha=\tan ^{-1} \frac{v _y}{v_ x}=\tan ^{-1} \frac{-3.43}{1.1}=-72.2^{\circ}$

$\begin{array}{l}{\text { A grasshoper leaps into the air from the edge of a vertical cliff, as shown in Fig. }} \\ {P 3.63 \text { . Use information from the figure to find (a) the initial speed of the }} \\ {\text { grasshoper and (b) the height of the cliff. }}\end{array}$

$y-y_{0}=6.74 \mathrm{cm} \longrightarrow 0.0674 \mathrm{m}$

$v_{y}=0, \quad a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$

${v_{ y}}^{2}={v_{0 y}}^{2}+2 a_ y\left(y-y_{0}\right)\rightarrow V_{0 y}=\sqrt{2 {a_y}\left(y-y_{0}\right)}$

$=\sqrt{2(-9.8)(0.0674)}=1.15 \mathrm{m} / \mathrm{s}$

$V_\text { 0y } \rightarrow \text V_0 ? ? \rightarrow V_{\text {0y }}=V_{0} \sin \alpha_{0}$

$V_{0}=\frac{V_{0y}}{\sin \alpha}=\frac{1.15}{\sin (50)}=1.5 \mathrm{m} / \mathrm{s}$

$x-x_{0}=1.06 \mathrm{m}$

$x-x_{0}=V_{0x}t+\frac{1}{2} a_ x t^2$

$V_{0 x}=V_{0} \cos \alpha \longrightarrow t= \frac{x-x_{0}}{V_{0 x}}=\frac{1.06}{1.5 \cos 50}=1.1 \mathrm{s}$

$t=1.1 s$

$y-y_{0}=6.74 cm \longrightarrow 0.0674 m \quad, \quad x-x_{0}=1.06m \quad, t=1.1s$

$v_{y}=0, \quad a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$

$y-y_{0}=v_{0 y} t+\frac{1}{2} a_ y t^{2}$

$=(1.15)(1.1)+\frac{1}{2}(-9.8)(1.1)^{2}=-4.66 \mathrm{m}$