Author Archives: Engineering Math

College Physics by Openstax Chapter 2 Problem 48

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

Solution:

The known values are: y0=2.20 my_0=2.20\:\text{m}; y=1.80 my=1.80\:\text{m}; v0=11.0 m/sv_0=11.0\:\text{m/s}; and a=−9.80 m/s2a=-9.80\:\text{m/s}^2

We are going to use the formula

Δy=v0t+12at2 \Delta y=v_0t+\frac{1}{2}at^2

Substituting the given values:

Δy=v0t+12at21.80 m−2.20 m=(11.0 m/s)t+12(−9.80 m/s2)t2−0.40 m=(11.0 m/s)t−(4.90 m/s2)t24.90t2−11t−0.40=0\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ 1.80\:\text{m}-2.20\:\text{m} & =\left(11.0\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\ -0.40\:\text{m} & =\left(11.0\:\text{m/s}\right)t-\left(4.90\:\text{m/s}^2\right)t^2 \\ 4.90t^2-11t-0.40 & =0 \end{align*}

Using the quadratic formula solve for tt, we have

t=−(−11)±(−11)2−4(4.90)(−0.40)2(4.90)\begin{align*} t & =\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)} \\ \end{align*}
t=2.28 sec     or      t=−0.04 sec t=2.28\:\text{sec}\:\:\:\:\:\text{or}\:\:\:\:\:\:t=-0.04 \ \text{sec}

We can discard the negative time, so

t=2.28 s  (Answer)t=2.28\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 2 Problem 47

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s.

(b) How long would it take to reach the ground if it is thrown straight down with the same speed?

Solution:

Part A

Refer to the figure below.

The known values are: t=2.35 s t=2.35\:\text{s}; y=0 my=0\:\text{m}; v0=+8.00 m/sv_0=+8.00\:\text{m/s}; and a=−9.8 m/s2a=-9.8\:\text{m/s}^2

Based on the given values, the formula that we shall use is

y=y0+v0t+12at2y=y_0+v_0t+\frac{1}{2}at^2

Substituting the values, we have

y=y0+v0t+12at20 =y0+(8.00 m/s)(2.35 s)+12(−9.80 m/s2)(2.35 s)2y0=8.26 m  (Answer)\begin{align*} y & =y_0+v_0t+\frac{1}{2}at^2 \\ 0\: & =y_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2 \\ y_0 & =8.26\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Therefore, the cliff is 8.26 meters high.

Part B

Refer to the figure below

The knowns now are: y=0 my=0\:\text{m}; y0=8.26 my_0=8.26\:\text{m}; v0=−8.00 m/sv_0=-8.00\:\text{m/s}; and a=−9.80 m/s2a=-9.80\:\text{m/s}^2

Based on the given values, we can use the formula

y=y0+v0t+12at2y=y_0+v_0t+\frac{1}{2}at^2

Substituting the values, we have

y=y0+v0t+12at20 m=8.26 m+(−8.00 m/s)t+12(−9.80 m/s2)t24.9t2+8t−8.26=0\begin{align*} y & =y_0+v_0t+\frac{1}{2}at^2 \\ 0\:\text{m} & =8.26\:\text{m}+\left(-8.00\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\ 4.9 t^2+8t-8.26 & =0 \\ \end{align*}

Using the quadratic formula to solve for the value of tt, we have

t=−8±(8)2−4(4.9)(−8.26)2(4.9)t=0.717 s  (Answer)\begin{align*} t &=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)} \\ t &=0.717\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 46

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.

(a) How long are her feet in the air?

(b) What is her highest point above the board?

(c) What is her velocity when her feet hit the water?

Solution:

The known values are: y0=1.80 my_0=1.80\:\text{m}, y=0 my=0\:\text{m}, a=−9.80 m/s2a=-9.80\:\text{m/s}^2, v0=4.00 m/sv_0=4.00\:\text{m/s}.

Part A

Based from the knowns, the formula most applicable to solve for the time is Δy=v0t+12at2\Delta y=v_0t+\frac{1}{2}at^2. If we rearrange the formula by solving for tt, and substitute the given values, we have

t=−v0±v02−2aΔyat=−4.00 m/s±(4.00 m/s)2−2(−9.80 m/s2)(1.80 m)−9.80 m/s2t=1.14 s  (Answer)\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a} \\ t & =\frac{-4.00\:\text{m/s}\pm \sqrt{\left(4.00\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.80\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\ t & =1.14\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We have the formula

Δy=v2−v022aΔy=(0 m/s)2−(4.00 m/s)22(−9.80 m/s2)Δy=0.816 m  (Answer)\begin{align*} \Delta y & =\frac{v^2-v_0^2}{2a} \\ \Delta y & =\frac{\left(0\:\text{m/s}\right)^2-\left(4.00\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)} \\ \Delta y & =0.816\: \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The formula to be used is

v2=v02+2aΔyv^2=v_0^2+2a\Delta y

Substituting the given values

v2=v02+2aΔyv=±v02+2aΔyv=±(4.00 m/s)2+2(−9.80 m/s2)(−1.80 m)v=±51.28 m2/s2v=±7.16 m/s\begin{align*} v^2 & =v_0^2+2a\Delta y \\ v & =\pm \sqrt{v_0^2+2a\Delta y} \\ v & =\pm \sqrt{\left(4.00\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-1.80\:\text{m}\right)} \\ v & =\pm \sqrt{51.28\:\text{m}^2/\text{s}^2} \\ v &=\pm 7.16\:\text{m/s} \end{align*}

Since the diver must be moving in the negative direction,

v=−7.16 m/s  (Answer)v=-7.16\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 2 Problem 45

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.(a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable.(c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.

Solution:

We will treat the downward direction as negative, and the upward direction as positive.

Part A

The known values are:a=−9.80 m/s2a=-9.80\:\text{m/s}^2; v0=13 m/sv_0=13\:\text{m/s}; and y0=0 my_0=0\:\text{m}.

Part B

At the highest point of the jump, the velocity is equal to 00. For this part, we will treat the initial position at the moment it jumps out of the water, and the final position at the highest point. Therefore, vf=0m/sv_f=0 \text{m/s}.

The unknown is the final position, yfy_f. We are going to use the formula

(vf)2=(v0)2+2aΔyor(vf)2=(v0)2+2a(yf−y0)\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta y \\ \text{or} \\ \left(v_f\right)^2=\left(v_0\right)^2+2a\left(y_f-y_0\right)

Solving for yfy_f in terms of the other variables:

yf=(vf)2−(v0)22a+y0y_f=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}+y_0

Substituting the given values:

yf=(vf)2−(v0)22a+y0yf=(0 m/s)2−(13.0 m/s)22(−9.80 m/s2)+0 myf=8.62 m+0 myf=8.62 m  (Answer)\begin{align*} y_f & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}+y_0 \\ y_f & =\frac{\left(0\:\text{m/s}\right)^2-\left(13.0\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)}+0\:\text{m} \\ y_f & =8.62\:\text{m}+0\:\text{m} \\ y_f & =8.62\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This value is reasonable since dolphins can jump several meters high out of the water. Usually, a dolphin measures about 2 meters and they can jump several times their length.

Part C

The unknown is time, Δt\Delta t. We are going to use the formula

vf=v0+atv_f=v_0+at

Solving for time, Δt\Delta t in terms of the other variables:

t=vf−v0at=\frac{v_f-v_0}{a}

Substituting the given values:

t=vf−v0at=0 m/s−13.0 m/s−9.80 m/s2t=1.3625 s\begin{align*} t & =\frac{v_f-v_0}{a} \\ t & =\frac{0\:\text{m/s}-13.0\:\text{m/s}}{-9.80\:\text{m/s}^2} \\ t &=1.3625\:\text{s} \end{align*}

This value is the time it takes the dolphin to reach the highest point. Since the time it takes to reach this point is equal to the time it takes to go back to the water, the time it is in the air is:

tair=2×ttair=2×1.3625 stair=2.65 s  (Answer)\begin{align*} t_{air} & =2\times t \\ t_{air}&=2\times 1.3625\:\text{s} \\ t_{air}&=2.65\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 44

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.

Solution:

We will treat the upward direction as positive, and the downward direction as negative.

Part A

The known values are: a=−9.80m/s2a=-9.80 \text{m/s}^2; v0=−1.40 m/sv_0=-1.40\:\text{m/s}; Δt=1.8 s\Delta t=1.8\:\text{s}; and yf=0 my_f=0\:\text{m}

Part B

We are looking for the initial position, y0y_0. We are going to use the formula

Δy=v0yt+12at2oryf−y0=v0yt+12at2\Delta y=v_{0y}t+\frac{1}{2}at^2 \\ \text{or} \\ y_f-y_0=v_{0y}t+\frac{1}{2}at^2

Solving for y0y_0 in terms of the other variables:

y0=yf−v0yt−12at2y_0=y_f-v_{0y}t-\frac{1}{2}at^2

Substituting the given values:

y0=yf−v0yt−12at2y0=0−(−1.4 m/s)(1.8 s)−12(−9.80 m/s2)(1.8 s)2y0=0−(−1.4 m/s)(1.8 s)−12(−9.80 m/s2)(1.8 s)2y0=0+2.52 m+15.876 my0=18.396 m  (Answer)\begin{align*} y_0 & =y_f-v_{0y}t-\frac{1}{2}at^2 \\ y_0& =0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\ y_0&= 0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\ y_0& = 0+2.52\:\text{m}+15.876\:\text{m} \\ y_0& =18.396\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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College Physics by Openstax Chapter 2 Problem 43

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

Solution:

It is our assumption that the player attempts to get the ball at the top where the velocity is zero.

The given are the following: vfy=0 m/sv_{fy}=0 \ \text{m/s}; Δy=1.25 m\Delta y=1.25 \ \text{m}; and a=−9.80 m/s2a=-9.80 \ \text{m/s}^2.

We are required to solve for the initial velocity v0yv_{0y} of the player. We are going to use the formula

(vfy)2=(voy)2+2aΔy\left(v_{fy}\right)^2=\left(v_{oy}\right)^2+2a\Delta y

Solving for voyv_{oy} in terms of the other variables:

voy=(vfy)2−2aΔyv_{oy}=\sqrt{\left(v_{fy}\right)^2-2a\Delta y}

Substituting the given values:

voy=(vfy)2−2aΔyvoy=(0 m/s)2−2(−9.80 m/s2)(1.25 m)voy=4.95 m/s  (Answer)\begin{align*} v_{oy} & =\sqrt{\left(v_{fy}\right)^2-2a\Delta y} \\ v_{oy} & = \sqrt{\left(0\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.25\:\text{m}\right)} \\ v_{oy} & =4.95 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 42

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Solution:

The given known quantities are: a=−9.8 m/s2a=-9.8\:\text{m/s}^2; y0=0 my_0=0 \ \text{m}; and v0=−14 m/sv_0=-14 \ \text{m/s}.

To compute for the displacement, we use the formula

Δy=v0t+12at2\Delta y=v_0t+\frac{1}{2}at^2

and to compute for the final velocity, we use the formula

vf=v0+atv_f=v_0+at

Part A

The displacement at t=0.500 st=0.500 \ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(0.500 s)+12(−9.8 m/s2)(0.500 s)2Δy=−8.23 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\ \Delta y & =-8.23\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=0.500 st=0.500 \ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(0.500 s)=−18.9 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\ & =-18.9\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The displacement at t=1.00 st=1.00\ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(1.00 s)+12(−9.8 m/s2)(1.00 s)2Δy=−18.9 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right)^2 \\ \Delta y & =-18.9\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.00 st=1.00\ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(1.00 s)=−23.8 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right) \\ & =-23.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The displacement at t=1.50 st=1.50\ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(1.50 s)+12(−9.8 m/s2)(1.50 s)2Δy=−32.0 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right)^2 \\ \Delta y & =-32.0\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.50 st=1.50\ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(1.50 s)=−28.7 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right) \\ & =-28.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The displacement at t=2.00 st=2.00\ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(2.00 s)+12(−9.8 m/s2)(2.00 s)2Δy=−47.6 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right)^2 \\ \Delta y & =-47.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=2.00 st= 2.00 \ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(2.00 s)=−33.6 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right) \\ & =-33.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part E

The displacement at t=2.50 st=2.50\ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(2.50 s)+12(−9.8 m/s2)(2.50 s)2Δy=−65.6 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right)^2 \\ \Delta y & =-65.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=2.50 st= 2.50 \ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(2.50 s)=−38.5 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right) \\ & =-38.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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