To solve for the final velocity, remember that the relationship between velocity, acceleration, and time is \(\displaystyle v_f=v_i+at\).

Using the given values for the initial velocity, acceleration, and time, we can solve for the final velocity.

\(\displaystyle v_f=1.5\frac{m}{s}+(2\frac{m}{s^2}*5.23s)\)

\(\displaystyle v_f=1.5\frac{m}{s}+10.46\frac{m}{s}\)

\(\displaystyle v_f=11.96\frac{m}{s}\)

We can solve this question using the equation for acceleration in terms of velocity:

\(\displaystyle a=\frac{v_2-v_1}{t}\)

We know our initial velocity (zero, since we start from rest), time, and the acceleration of gravity. Use these values to isolate the variable for the final velocity.

\(\displaystyle -9.8\frac{m}{s^2}=\frac{v_2-0\frac{m}{s}}{2.1s}\)

\(\displaystyle (-9.8\frac{m}{s^2})(2.1s)=v_2\)

\(\displaystyle -20.58\frac{m}{s}=v_2\)

Note that the final velocity is negative, since the book is traveling in the downward direction.

Use the kinematic equation:

\(\displaystyle v_{f} = v_{i} + at\) 

We are given the initial velocity, the time elapsed, and the acceleration. Using these values, we can solve for the final velocity.

\(\displaystyle v_f=25\frac{m}{s}+(2\frac{m}{s^2})(15s)\)

\(\displaystyle v_f=25\frac{m}{s}+30\frac{m}{s}\)

\(\displaystyle v_f=55\frac{m}{s}\)

Utilize the kinematic equation:

\(\displaystyle x = v_{o}t + \frac{1}{2}at^{2}\)

The particle's motion can be broken into two parts: the initial distance and the distance traveled during acceleration. The initial distance is given.

\(\displaystyle x_o=35000m\)

The distance during acceleration can be found using the kinematic formula and given values for the initial velocity, acceleration, and time.

\(\displaystyle x_a = (850\frac{m}{s})(15s) + \frac{1}{2}(12\frac{m}{s^2})(15s)^{2}\)

\(\displaystyle x_a=(12750m)+(1350m)\)

\(\displaystyle x_a=14100m\)

Add the two distances together.

\(\displaystyle x_o+x_a=35000m+14100m=49100m\)

Convert the final answer to kilometers.

\(\displaystyle 49100m*\frac{1km}{1000m}=49.1km\)

Although this question involves several steps, when you break it down we can see that it is a problem involving kinematics in two dimensions.

First, begin by writing down what we know:

\(\displaystyle t=3.45s\)

Distance to goal post: \(\displaystyle x_1=58.5m\)

Distance past goal post: \(\displaystyle x_2=4.50m\)

What we want to know:

\(\displaystyle v_0=?\)

\(\displaystyle \theta=?\)

To begin, we need to calculate each of the vector components of the velocity. We can start with the horizontal component. We know that velocity is equal to distance divided by time, and that horizontal velocity, \(\displaystyle v_x\), will not change because there is no acceleration in the horizontal direction. We will need to find the total distance that the ball travels in order to solve.

\(\displaystyle v_x=\frac{x}{t}\)

We will need to find the total distance that the ball travels in order to solve.

\(\displaystyle x=x_1+x_2\)

\(\displaystyle x=58.5m+4.50m\)

\(\displaystyle x=63.0m\)

Use this distance and the given time to find the horizontal velocity.

\(\displaystyle v_x=\frac{63.0m}{3.45s}\)

\(\displaystyle v_x=18.3\frac{m}{s}\)

Now let's find the initial vertical velocity,\(\displaystyle v_{oy}\). Because we are assuming that there is nothing except for gravity influencing the ball, we can say that the ball spends half of the time reaching the peak of its trajectory, where the vertical velocity will momentarily be zero. With that information, we can solve for the initial vertical velocity:

\(\displaystyle v_y=v_{oy}+a(\frac{1}{2}t)\)

We use only half the given time because we are only taking the time from when the ball is kicked to when it reaches the top of its trajectory (which will be half of its flight). As we stated above, velocity will be zero at the top of the trajectory. We are using a negative value for the acceleration due to gravity because gravity points downward, which in this case is the negative direction. Use the given values to solve for the initial vertical velocity.

\(\displaystyle v_{oy}=v_y-a(\frac{1}{2}t)\)

\(\displaystyle v_{oy}=0\frac{m}{s}-(-9.81\frac{m}{s^2})(\frac{1}{2}*3.45s)\)

\(\displaystyle v_{oy}=16.9\frac{m}{s}\)

Now that we have both directional components of the initial velocity, we can use the Pythagorean theorem to solve for the total initial velocity.

\(\displaystyle {v_o}^2={v_x}^2+{v_o_y}^2\)

\(\displaystyle v_o=\sqrt{{v_x}^2+{v_o_y}^2}\)

\(\displaystyle v_o=\sqrt{(18.3\frac{m}{s})^2+(16.9\frac{m}{s})^2}\)

\(\displaystyle v_o=24.9\frac{m}{s}\)

Now to find the angle, we use trigonometry. In a triangle formed by the maximum height of the ball, the ground, and the trajectory, tangent of the angle will be equal to the vertical leg of the triangle divided by the horizontal leg of the triangle.

\(\displaystyle tan\theta=\frac{v_{oy}}{v_x}\)

\(\displaystyle \theta=tan^{-1}(\frac{v_{oy}}{v_x})\)

\(\displaystyle \theta=tan^{-1}(\frac{16.9}{18.3})\)

\(\displaystyle \theta=42.8^0\)

To solve this problem, we need to calculate each of the vector components of the velocity. Our plan is to find the time it takes for the ball to horizontally reach the goal post. Once we have that time, we can apply that time to the vertical component to see at what height the ball is at this point. We know that velocity is equal to distance divided by time, and that horizontal velocity, \(\displaystyle v_x\), will not change because there is no acceleration in the horizontal direction. We will need to find the total distance that the ball travels in order to solve.

\(\displaystyle v_x=\frac{x}{t}\)

\(\displaystyle x=x_1+x_2\)

\(\displaystyle x=58.5m+4.50m\)

\(\displaystyle x=63.0m\)

Use this distance and the given time to find the horizontal velocity.

\(\displaystyle v_x=\frac{63.0m}{3.45s}\)

\(\displaystyle v_x=18.3\frac{m}{s}\)

Now let's find the initial vertical velocity,\(\displaystyle v_{oy}\). Because we are assuming that there is nothing except for gravity influencing the ball, we can say that the ball spends half of the time reaching the peak of its trajectory, where the vertical velocity will momentarily be zero. With that information, we can solve for the initial vertical velocity:

\(\displaystyle v_y=v_{oy}+a(\frac{1}{2}t)\)

We use only half the given time because we are only taking the time from when the ball is kicked to when it reaches the top of its trajectory (which will be half of its flight). As we stated above, velocity will be zero at the top of the trajectory. We are using a negative value for the acceleration due to gravity because gravity points downward, which in this case is the negative direction. Use the given values to solve for the initial vertical velocity.

\(\displaystyle v_{oy}=v_y-a(\frac{1}{2}t)\)

\(\displaystyle v_{oy}=0\frac{m}{s}-(-9.81\frac{m}{s^2})(\frac{1}{2}*3.45s)\)

\(\displaystyle v_{oy}=16.9\frac{m}{s}\)

Now that we have our velocities, let's see how long it takes for the ball to reach the goal post. We will use \(\displaystyle t_p\) as that time.

\(\displaystyle v_x=\frac{x_1}{t_p}\)

\(\displaystyle t_p=\frac{x_1}{v_x}\)

\(\displaystyle t_p=\frac{58.5m}{13.8\frac{m}{s}}\)

\(\displaystyle t_p=3.20s\)

Now that we have our time, let's use a different equation to see how high the ball is at that time. We know the time, initial vertical velocity, and acceleration. Use these to find the final vertical distance, or height.

\(\displaystyle y=v_{oy}t_p+\frac{1}{2}a{t_p}^2\)

\(\displaystyle y=(16.9\frac{m}{s})(3.20s)+\frac{1}{2}(-9.81\frac{m}{s^2})(3.20s)^2\)

\(\displaystyle y=3.85m\)

This gives us the height of the ball when it is just over the goal post. We know from the question that there is a gap of \(\displaystyle 80cm\) between the ball and the height of the goal post. To find that height, we take the ball's height at this moment and subtract the height of the gap.

\(\displaystyle h=3.85m-0.80m\)

\(\displaystyle h=3.05m\)

First we need to know how long this chase takes. The dog doesn't help us there, but we know that the car is going along at a constant velocity and will therefore cover the same distance in the same amount of time.

Since the velocity is constant, we can say \(\displaystyle \Delta x =v*t\).

Plug in our given values.

\(\displaystyle \Delta x =v*t\)

\(\displaystyle 7m=13\frac{m}{s}*t\)

\(\displaystyle \frac{7m}{13\frac{m}{s}}=t\)

\(\displaystyle 0.54\: s=t\)

Plug that into our equation for the dog. We can use the full distance equation for this: \(\displaystyle \Delta x=v_it+\frac{1}{2}at^2\)

Remember the dog starts at rest.

Plug in our given information.

\(\displaystyle \Delta x=v_it+\frac{1}{2}at^2\)

\(\displaystyle 7m=0\frac{m}{s}*0.54s+\frac{1}{2}a(0.54s)^2\)

\(\displaystyle 7m=\frac{1}{2}a(0.54s)^2\)

\(\displaystyle 7m=\frac{1}{2}a*0.2916s^2\)

\(\displaystyle 7m=0.1458s^2*a\)

\(\displaystyle \frac{7m}{0.1458s^2}=a\)

\(\displaystyle 48\frac{m}{s^2}=a\)

To solve this problem we will need to use the appropriate kinematics equation:

\(\displaystyle v_f^2=v_i^2+2a\Delta x\)

Remember that when an object is thrown vertically, it will have a velocity of zero at the highest point. We can use this as our final velocity. The initial velocity is given, and the acceleration will be the acceleration due to gravity. Using these values, we can solve for the displacement when the ball is at its peak.

\(\displaystyle v_f^2=v_i^2+2a\Delta x\)

\(\displaystyle (0\frac{m}{s})^2=(0.44\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta x\)

\(\displaystyle 0\frac{m^2}{s^2}=(0.1936\frac{m^2}{s^2})+(-19.6\frac{m}{s^2})\Delta x\)

\(\displaystyle -0.1936\frac{m^2}{s^2}=(-19.6\frac{m}{s^2})\Delta x\)

\(\displaystyle \frac{-0.1936\frac{m^2}{s^2}}{-19.6\frac{m}{s^2}}=\Delta x\)

\(\displaystyle 0.01m=\Delta x\)

Since there is only one force acting upon the object, the force due to friction, we can find its value using the equation \(\displaystyle F=ma\). The problem gives us the mass of the crate, but we have to solve for the acceleration.

Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

\(\displaystyle \Delta x=\frac{(v_f+v_i)}{2}\Delta t\)

Plug in our given values and solve.

\(\displaystyle 10m=\frac{(0\frac{m}{s}+v_i)}{2}5s\)

\(\displaystyle \frac{10m}{5s}=\frac{v_i}{2}\)

\(\displaystyle 2\frac{m}{s}=\frac{v_i}{2}\)

\(\displaystyle 4\frac{m}{s}=v_i\)

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

\(\displaystyle \Delta x=v_it+\frac{1}{2}at^2\)

Plug in the given values to solve for acceleration.

\(\displaystyle 10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2\)

\(\displaystyle 10m=(20m)+\frac{1}{2}a(25s^2)\)

\(\displaystyle -10m=\frac{1}{2}a(25s^2)\)

\(\displaystyle -20m=a(25s^2)\)

\(\displaystyle \frac{-20m}{25s^2}=a\)

\(\displaystyle -0.8\frac{m}{s^2}=a\)

Now that we have the acceleration and the mass, we can solve for the force of friction.

\(\displaystyle F=ma\)

\(\displaystyle F=(2.9kg)(-0.8\frac{m}{s^2})\)

\(\displaystyle F=-2.32N\) 

The important thing to recognize here is that the amount of time the egg is falling will be equal to the amount of time the girl is running.

Our first step will be to find the time that the egg is in the air.

We know it starts from rest \(\displaystyle 2.8m\) above the ground, and we know the gravitational acceleration. Its total displacement will be \(\displaystyle -2.8m\), since it falls in the downward direction. We can use the appropriate motion equation to solve for the time:

\(\displaystyle \Delta y =v_it+\frac{1}{2}at^2\)

Use the given values in the formula to solve for the time.

\(\displaystyle -2.8m=(0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -2.8m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -2.8m=(-4.9\frac{m}{s^2})t^2\)

\(\displaystyle \frac{-2.8m}{-4.9\frac{m}{s^2}}=t^2\)

\(\displaystyle 0.57s^2=t^2\)

\(\displaystyle 0.755s=t\)

Now that we have the time, we can use it to find the speed of the girl. Her speed will be determined by the distance she travels in this amount of time.

\(\displaystyle v=\frac{d}{t}\)

Use our values for her distance and the time to solve for her velocity.

\(\displaystyle v=\frac{13m}{0.755s}\)

\(\displaystyle v=17.22\frac{m}{s}\)