Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?
Solution:
Part A
We know that the initial height, y 0 y_0 y 0 v 0 v_0 v 0
The change in height is
Δ y = v 0 t + 1 2 at 2 = ( 0 ) ( 1.50 s ) + 1 2 ( 9.81 m/s 2 ) ( 1.50 s ) 2 = 0 + 11.036 m = 11.04 m \displaystyle \begin{aligned}
\Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\
&=\left( 0 \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\
&=0+11.036\ \text{m} \\
&=11.04 \ \text{m}
\end{aligned} Δ y = v 0 t + 2 1 at 2 = ( 0 ) ( 1.50 s ) + 2 1 ( 9.81 m/s 2 ) ( 1.50 s ) 2 = 0 + 11.036 m = 11.04 m So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is
y = y 0 − Δ y = 105 m − 11.04 m = 93.96 m \displaystyle \begin{aligned}
\text{y}&=\text{y}_{0}-\Delta \text{y} \\
&=105\ \text{m}-11.04\ \text{m} \\
&=93.96 \ \text{m}
\end{aligned} y = y 0 − Δ y = 105 m − 11.04 m = 93.96 m Part B
We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is
y = 1 2 at 2 Solving for t, we have t = 2y a = 2 ( 105 m ) 9.81 m/s 2 = 4.63 s \begin{aligned}
\text{y} & =\frac{1}{2}\text{at}^{2} \\
&\text {Solving for t, we have}\\
\text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\
&=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\
&=4.63 \ \text{s}
\end{aligned}
y t = 2 1 at 2 Solving for t, we have = a 2y = 9.81 m/s 2 2 ( 105 m ) = 4.63 s Therefore, to move out the hiker has about
t = 4.63 s − 1.50 s = 3.13 s \begin{aligned}
\text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\
&=3.13\ \text{s}
\end{aligned}
t = 4.63 s − 1.50 s = 3.13 s
This entry was posted in Engineering Mathematics Blog , Physics , Sciences and tagged College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Freefall , motion of falling rocks , Physics , Physics Solutions , Solution Manual for College Physics by Openstax , the time for hikers to move out on January 5, 2021 by Engineering Math .
A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?
Part A
The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.
From
v 2 = ( v 0 ) 2 + 2 ay , \begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\
\end{aligned} v 2 = ( v 0 ) 2 + 2 ay , we have
v 2 = ( v 0 ) 2 + 2 ay v 2 − 2 ay = ( v 0 ) 2 v 0 = v 2 − 2 ay \begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\
\text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}}
\end{aligned} v 2 v 2 − 2 ay v 0 = ( v 0 ) 2 + 2 ay = ( v 0 ) 2 = v 2 − 2 ay Substituting the known values,
v 0 = v 2 − 2 ay v 0 = 0 2 − 2 ( − 9.81 m/s 2 ) ( 2.50 m ) v 0 = 7.00 m/s \begin{aligned}
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\
\text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\
\text{v}_0&= {\color{green}7.00 \ \text{m/s}}
\end{aligned} v 0 v 0 v 0 = v 2 − 2 ay = 0 2 − 2 ( − 9.81 m/s 2 ) ( 2.50 m ) = 7.00 m/s Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.
Part B
Since the motion of the kangaroo has uniform acceleration, we can use the formula
y = v o t + 1 2 a t 2 \text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2 y = v o t + 2 1 a t 2 The initial and final position of the kangaroo will be the same, so y y y 2 .
y = v 0 t + 1 2 a t 2 0 = ( 7.00 m/s ) t + 1 2 ( − 9.81 m/s 2 ) t 2 0 = 7 t − 4.905 t 2 7 t − 4.905 t 2 = 0 t ( 7 − 4.905 t ) = 0 t = 0 or 7 − 4.905 t = 0 \begin{aligned}
\text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\
0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\
0 & =7\text{t}-4.905\text{t}^{2}\\
7\text{t}-4.905\text{t}^{2}&=0 \\
\text{t}\left( 7-4.905\text{t} \right) & =0 \\
\text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\
\end{aligned} y 0 0 7 t − 4.905 t 2 t ( 7 − 4.905 t ) t = 0 = v 0 t + 2 1 a t 2 = ( 7.00 m/s ) t + 2 1 ( − 9.81 m/s 2 ) t 2 = 7 t − 4.905 t 2 = 0 = 0 or 7 − 4.905 t = 0 Discard the time 0 since this refers to the beginning of motion. Therefore, we have
7 − 4.905 t = 0 4.905 t = 7 t = 7 4.905 t = 1.43 s \begin{aligned}
7-4.905\text{t} &=0 \\
4.905\text{t} & = 7 \\
\text{t} & =\frac{7}{4.905} \\
\text{t}&={\color{green}1.43 \ \text{s}}
\end{aligned} 7 − 4.905 t 4.905 t t t = 0 = 7 = 4.905 7 = 1.43 s The kangaroo is about 1.43 seconds long in the air.
This entry was posted in Physics , Sciences and tagged College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Freefall , motion , motion of a kangaroo , Physics , Physics Solutions , Solution Manual for College Physics by Openstax on December 26, 2020 by Engineering Math .
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This entry was posted in Physics , Sciences and tagged College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Conservation of Energy , Conservative Forces , GPE , gravitational potential energy , kinetic energy , Nonconservative Forces , Physics , Potential Energy , Power , work , Work Energy and Power in Humans , work-energy , work-energy theorem on November 1, 2020 by Engineering Math .
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This entry was posted in Physics , Sciences and tagged Angular Velocity , centripetal acceleration , centripetal force , College Physics , College Physics by Openstax , College Physics by Openstax Solution Manual , College Physics Complete Solution Manual , College Physics Solutions , Force and Motion , Newton's Laws of Motion , Physics , Rotation Angle on November 1, 2020 by Engineering Math .
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