Using the equation \(\displaystyle d=v_ot + (1/2)a(t^2)\) we can find the distance at which the ball was dropped. Notice that the mass of the ball does not matter in this problem. We are told that the ball is dropped from rest making, \(\displaystyle v_o =0\), thus we have \(\displaystyle d=(1/2)a(t^2)\). When we plug in our values, and assuming that acceleration is equal to gravity (10m/s²) we find that  \(\displaystyle d=(1/2)(10)(5^2)\) = 125m.

Since the object is dropped, the inital velocity is zero. Gravity is the only acceleration, the time is ten seconds, and the distance at which the object travels is unknown.

The equation \(\displaystyle d=v_ot +(1/2)at^2\) can be used to find the distance traveled.

\(\displaystyle d=0+(1/2)(10m/s^2)(10s)^2=500m\)

Using the equation \(\displaystyle d=v_ot + (1/2)at^2\), we can solve for time.

Since the object started at rest, \(\displaystyle v_0=0\). Now we are left with the equation \(\displaystyle d=(1/2)at^2\).

\(\displaystyle 30m=\frac{1}{2}(2\frac{m}{s^2})t^2\)

\(\displaystyle 30s^2=t^2\)

\(\displaystyle t=5.5s\)

Plugging in the remaining values we can find that t = 5.5s.

Use the following kinematic equation: \(\displaystyle y=y_o+v_ot+\frac{1}{2}at^2\).

We can choose the ground to be the zero distance so that \(\displaystyle y=0\) and \(\displaystyle y_o=200\ \text{m}\).

Also, the initial speed is zero.

\(\displaystyle v_o=0\)

The kinematic equation simplifies using these values.

\(\displaystyle 0=y_o+\frac{1}{2}at^2\)

Solve to isolate the time variable.

\(\displaystyle t=\sqrt{\frac{-2y_o}{a}}\)

We know that the acceleration is the acceleration due to gravity. Now we can plug in the known values and solve.

\(\displaystyle t=\sqrt{\frac{-2(200\ \text{m})}{-9.8\ \frac{\text{m}}{\text{s}^2}}}\)

\(\displaystyle t=\sqrt{40.8s^2}=6.4s\)

Use the following kinematic equation where the initial velocity is \(\displaystyle v_o\), final velocity is \(\displaystyle v_f\), and the distance traveled is \(\displaystyle \Delta x\).

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

We can use the values in the question to solve for the acceleration.

\(\displaystyle (0\frac{m}{s})^2=(20\frac{m}{s})^2+2a(28m)\)

We rearrange the equation to solve for the acceleration.

\(\displaystyle a=\frac{0^2-(20\frac{m}{s})^2}{2(28\ \text{m})}\)

\(\displaystyle a=-7.14\ \frac{\text{m}}{\text{s}^2}\)

The first step to solving this problem will be to find the velocity of the ball at the point when the picture is taken. We know the initial velocity of the ball (zero), the displacement, and the acceleration. Using the appropriate kinematics formula, we can solve for the final velocity.

\(\displaystyle v_f^2=v_o^2+2a\Delta y\)

\(\displaystyle v_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(5m-20m)\)

\(\displaystyle v_f^2=294\frac{m^2}{s^2}\)

\(\displaystyle v_f=17.15\frac{m}{s}\)

Now that we know how fast the ball is traveling when the picture is taken, we can find the distance it travels while the shutter is open. This distance will become a motion blur, making the vertical diameter of the ball appear stretched.

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

\(\displaystyle d=(17.15\frac{m}{s})(0.005s)=0.086m\)

During the exposure period, the ball will travel \(\displaystyle 0.086m\), or \(\displaystyle 8.6cm\). The diameter of the ball in the vertical direction will appear to be distorted by this distance.

\(\displaystyle 15cm+8.6cm=23.6cm\)

This is projectile motion in the vertical direction only, subject to the equation of motion: \(\displaystyle (x-x_o) = v_ot + \frac{1}{2}at^2\).

For this discussion, one can define the downward direction as negative. For projectile motion, \(\displaystyle a = -10 \frac{m}{s^2}\) (gravitational acceleration, or \(\displaystyle g\)).

In this case, the ball ends up \(\displaystyle 20 m\) below where is started, so \(\displaystyle (x - x_o) = -20 m\).

The initial velocity, \(\displaystyle v_o\), is \(\displaystyle 5 \frac{m}{s}\) (upward, thus positive).

With all this, the projectile motion equation becomes:

\(\displaystyle - 20 m = (5 \frac{m}{s})t - \frac{1}{2}(10 \frac{m}{s^2})t^2\)

This can be solved for \(\displaystyle t\) using the quadratic formula:

 \(\displaystyle t = \frac{-b \pm \sqrt{b^{2}-4ac)}) }{2a}\)

 The result is:

\(\displaystyle t = \frac{5\pm 20.62}{10} = 2.562 s\) (or \(\displaystyle -1.562 s\)).

Only the positive answer option is physically possible, and is thus our correct answer.

First we will find the time required for the ball to reach the ground. Since the ball is thrown horizontally, it has no initial vertical component. We use the following equation to solve for the total flight time:

\(\displaystyle \Delta y=v_{iy}t+\frac{1}{2}at^{2}\)

We are given the change in height, initial velocity, and acceleration. Using these values, we can solve for the time. Note that the change in height will be negative, since the ball is traveling downward.

\(\displaystyle -50m= (0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -50m=(-4.9\frac{m}{s^2})t^2\)

\(\displaystyle t=\sqrt{\frac{-50m}{-4.9\frac{m}{s^2}}}\)

\(\displaystyle t=3.19s\)

Finally, we use the horizontal velocity to find the distance traveled in \(\displaystyle 3.19s\). Remember that the horizontal velocity remains constant during projectile motion.

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

\(\displaystyle d=v*t=(15\frac{m}{s})*(3.19s)=47.9m\)

Since we know the change in velocity of the puck, we can determine the work done by friction by using the work-energy theorem:

\(\displaystyle W =Fd= \Delta KE\)

The work in this question is done by friction, so we can write:

\(\displaystyle F_f d = \Delta KE\)

Substituting in expressions for friction and kinetic energy, we get:

\(\displaystyle \mu_kF_Nd = \frac{1}{2}m(\Delta v^2)\)

The normal force is from gravity, so we can write:

\(\displaystyle \mu_kmgd=\frac{1}{2}m(\Delta v^2)\)

Rearranging for the coefficient of kinetic friction we get:

\(\displaystyle \mu_k = \frac{(\Delta v^2)}{2gd} = \frac{(2\frac{m}{s})^2}{2(10\frac{m}{s^2})(20m)} = 0.01\)

The equation needed is \(\displaystyle x = x_{0}+v_{0}t+\frac{1}{2}at^2\)

\(\displaystyle v_0 =0\)

\(\displaystyle 0=16m+\frac{1}{2}\left(-10\frac{m}{s^2}\right)t^2\)

\(\displaystyle t= 1.79 s\)