Solution:

## Grantham PHY220 Week 2 Problem 2

#### An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y axis and the total distance to the plane from you. Choose the x-axis east, y axis north, and z axis up.

Solution:

The angle of the plane in the x-y axis

Solve for $\theta$ $\theta =tan^{-1}\left(\frac{1500}{2100}\right)=35.5^{\circ}$

The angle is 35.5 degrees measured from the west axis. If it is measured from the East axis, the angle would be 180-35.5=144.5 degrees.

The distance from the plane to the observer

Solve for $d=\sqrt{\left(1500\:m\right)^2+\left(2100\:m\right)^2+\left(5000\:m\right)^2}=5626.72\:m$

## Solution:

The known values are $a=-9.80\:m/s^2;\:v_o=15.0\:m/s;\:y=7.00\:m$

The applicable formula is $y=v_ot+\frac{1}{2}at^2$.

Using this formula, we can solve it in terms of time, t. $t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}$

Substituting the known values, we have $t=\frac{-15.0\:m/s\pm \sqrt{\left(15.0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(7.00\:m\right)}}{-9.80\:m/s^2}$ $t=\frac{-15.0\:m/s\pm 9.37\:m/s}{-9.80\:m/s^2}$

We have two values for time, t. These two values represent the times when the ball passes the tree branch. So, the two values of the time, t, are $t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec$ $t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:sec$

Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds. That is $1.92 seconds$.

## Vector Addition and Subtraction| Two-Dimensional Kinematics| College Physics| Openstax| Problem 3.1|

#### Find the following for path A in Figure 3.54: (a) The total distance traveled, and (b) The magnitude and direction of the displacement from start to finish. Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

## Solution:

The known values are $\text{y}_0=2.20\:\text{m};\:\text{y}=1.80\:\text{m};\:\text{v}_0=11.0\:\text{m/s};\:\text{and}\:\text{a}=-9.80\:\text{m/s}^2$

The applicable formula is $\Delta y=v_0t+\frac{1}{2}at^2$. $1.80\:m-2.20\:m=\left(11.0\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2$ $-0.40\:m=\left(11.0\:m/s\right)t-\left(4.90\:m/s^2\right)t^2$ $4.90t^2-11t-0.40=0$

Using the quadratic formula solve for t, we have $\:t=\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)}$ $t=2.28\:sec\:\:\:\:\:or\:\:\:\:\:\:t=-0.04$

Therefore, $t=2.28\:s$

## Solution:

### Part A

Refer to the figure below.

The known values are: $\text{t}=2.35\:\text{s};\:\text{y}=0\:\text{m};\:\text{v}_0=+8.00\:\text{m/s};\:\text{a}=-9.8\:\text{m/s}^2$

Based from the given values, we can use the equation $y=y_0+v_0t+\frac{1}{2}at^2$. Substituting the values, we have $0\:=\text{y}_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2$ $y=8.26\:m$

Therefore, the cliff is 8.26 meters high.

### Part B

Refer to the figure below

The knowns now are: $y=0\:m;\:y_0=8.26\:m;\:v_0=-8.00\:m/s;\:a=-9.80\:m/s^2$

Based from the given values, we can use the equation $y=y_0+v_0t+\frac{1}{2}at^2$. Substituting the values, we have $0\:m=8.26\:m+\left(-8.00\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2$ $4.9\text{t}^2+8\text{t}-8.26=0$

Using the quadratic formula to solve for the value of t, we have $\displaystyle \text{t}=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)}$ $\text{t}=0.717\:\text{s}$

## Solution:

The known values are: $y_0=1.80\:m,\:y=0\:m,\:a=-9.80\:m/s^2,\:v_0=4.00\:m/s$

### Part A

Based from the knowns, the formula most applicable to solve for the time is $\Delta y=v_0t+\frac{1}{2}at^2$. If we rearrange the formula by solving for t, and substitute the given values, we have $t=\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a}$ $t=\frac{-4.00\:m/s\pm \sqrt{\left(4.00\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.80\:m\right)}}{-9.80\:m/s^2}$ $t=1.14\:s$

### Part B

We have the formula $\Delta y=\frac{v^2-v_0^2}{2a}$ $\Delta y=\frac{\left(0\:m/s\right)^2-\left(4.00\:m/s\right)^2}{2\left(-9.80\:m/s^2\right)}$ $\Delta y=0.816\:m$

### Part C

The formula to be used is $v^2=v_0^2+2a\Delta y$ $v=\pm \sqrt{v_0^2+2a\Delta y}$ $v=\pm \sqrt{\left(4.00\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(-1.80\:m\right)}$ $v=\pm \sqrt{51.28\:m^2/s^2}$ $v=\pm 7.16\:m/s$

Since the diver must be moving in the negative direction, $v=-7.16\:m/s$

## Solution:

### Part A

The knowns are $a=-9.80\:m/s^2$ $v_0=13\:m/s$ $y=0\:m$

### Part B

At the highest point of the jump, the velocity is equal to 0. That is $v=0\:m/s$

Basing from the known values, the formula that we can use to solve for y is $v^2=v_0^2+2a\Delta y$. By rearranging these variables, the formula in solving for $\Delta y$ is $\Delta y=\frac{v^2-v_0^2}{2a}$. Therefore we have, $\Delta y=\frac{\left(0\:m/s\right)^2-\left(13.0\:m/s\right)^2}{2\left(-9.80\:m/s^2\right)}$ $\Delta y=8.62\:m$

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 formula we can use to solve for the time is $v=v_0+at$. If we rearrange this formula and solve for t, it becomes $t=\frac{v-v_0}{a}$. $t=\frac{0\:m/s-13.0\:m/s}{-9.8\:m/s^2}$ $t=1.3625\:s$

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 water, the time it is in the air is $2.65\:s$.

## Solution:

### Part A

The knowns are $a=-9.80\:m/s^2$ $v_0=-1.40\:m/s$ $t=1.8\:s$ $y=0\:m$

### Part B $y_0=-y+v_0t+\frac{1}{2}at^2$ $y_0=0\:m\:+\left(-1.40\:m/s\right)\left(1.8\:s\right)+\frac{1}{2}\left(-9.80\:m/s^2\right)\left(1.8\:s\right)^2$ $y_0=18\:m$