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.

To find the final velocity we use the following kinematic equation.

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

The block starts from rest, so \(\displaystyle v_o=0\). The incline is eighteen meters high, so we can find the distance (hypotenuse) as follows:

\(\displaystyle \Delta x=\frac{18m}{sin24^\circ}\)

\(\displaystyle \Delta x = 44.25m\)

We know that the acceleration is \(\displaystyle a=2.2\frac{m}{s^2}\).

Plug in the values to find the velocity at the bottom of the incline.

\(\displaystyle v_f^2=(0\frac{m}{s})^2+2(2.2\frac{m}{s^2})(44.25m)\)

\(\displaystyle v_f=\sqrt{194.7\frac{m^2}{s^2}}=14\frac{m}{s}\)

The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, v_y, use \(\displaystyle v_y = v_o sin\theta\).

\(\displaystyle v_y = (75 \frac{m}{s})sin 45^o = 53 \frac{m}{s}\)

Next we find how long it takes to reach the top of its trajectory using \(\displaystyle v_f = v_o +at\).

\(\displaystyle 0 \frac{m}{s} = 53 \frac{m}{s} + (-10 \frac{m}{s^2})t\)

 t = 5.3s

Finally, find how high the object goes with \(\displaystyle d = v_ot +\frac{1}{2}at^{2}\).

\(\displaystyle d = (53\frac{m}{s})5.3s + \frac{1}{2}(-10\frac{m}{s^2})(5.3s)^{2} = 140 m\) 

First, find the horizontal (x) and vertical (y) components of the velocity

\(\displaystyle v_x = v_o cos\theta\)

\(\displaystyle v_x = (125 \frac{m}{s}) cos(30^o) = 108\frac{m}{s}\)

\(\displaystyle v_y = v_o sin\theta\)

\(\displaystyle v_y = (125\frac{m}{s})sin(30^o) = 62.5\frac{m}{s}\)

Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.

\(\displaystyle v_f = v_o + at\)

\(\displaystyle 0 \frac{m}{s} = 62.5 \frac{m}{s} + (-10 \frac{m}{s^2})t\)

t = 6.25s 

Total time in the air is therefore 12.5s (twice this value).

Finally, find distance traveled my multiplying horizontal velocity and time.

\(\displaystyle d_x = v_x(t)\)

\(\displaystyle d_x = 108 \frac{m}{s} (12.5s) = 1,350 m\)