Remember that at the highest point, the velocity in the y-direction is equal to zero. Using the given values and the equation below, we can solve for the initial velocity.

$\displaystyle V_f^2=V_i^2+2a\Delta y$

$\displaystyle (0\frac{m}{s})^2=V_i^2+2(-9.8\frac{m}{s^2})(10m)$

$\displaystyle 0\frac{m^2}{s^2}=V_i^2-196\frac{m^2}{s^2}$

$\displaystyle 196\frac{m^2}{s^2}=V_i^2$

$\displaystyle \sqrt{196\frac{m^2}{s^2}}=\sqrt{V_i^2}$

$\displaystyle 14\frac{m}{s}=V_i$

Remember that the final velocity at an object's highest point is equal to zero. We can use the following equation to solve for our height ($\displaystyle \Delta{y}}$), using the initial velocity, final velocity, and acceleration due to gravity.

$\displaystyle V_{f}^{2}=V_{i}^{2}+2a\Delta{y}}$

$\displaystyle (0\frac{m}{s})^2=({35\frac{m}{s}})^{2}+2(-9.8\frac{m}{s^2})\Delta{y}}$

$\displaystyle 0=1225\frac{m^2}{s^2}+(-19.6\frac{m}{s^2})\Delta{y}}$

$\displaystyle -1225\frac{m^2}{s^2}=(-19.6\frac{m}{s^2})\Delta{y}}$

$\displaystyle \frac{-1225\frac{m^2}{s^2}}{-19.6\frac{m}{s^2}} =\Delta{y}}$

$\displaystyle 62.5m=\Delta{y}$

The problem gives us the distance, the acceleration due to gravity, and implies that the initial velocity of the picture is zero, as it starts at rest.

We can find the final velocity using the appropriate motion equation:

$\displaystyle v_f^2=v_i^2+2a\Delta y$

We can use our values to solve for the final velocity. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

$\displaystyle v_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-1.5m)$

$\displaystyle v_f^2=29.4\frac{m^2}{s^2}$

$\displaystyle \sqrt{v_f^2}=\sqrt{29.4\frac{m^2}{s^2}}$

$\displaystyle v_f=5.42\frac{m}{s}$

The square root of a term can be either positive or negative, depending on the direction. Since velocity is a vector, and the painting is falling downward, our final answer will be negative: $\displaystyle -5.42\frac{m}{s}$.

The relationship between velocity, distance, and time is:

$\displaystyle v=\frac{\Delta x}{\Delta t}$

We can multiply both sides by the time, then divide both sides by the velocity, to isolate the variable for time.

$\displaystyle v*\Delta t=\Delta x$

$\displaystyle \Delta t=\frac{\Delta x}{v}$

The quotient of distance and velocity will give us the time that the object was in motion.

When examining vertical motion, the vertical velocity will always be zero at the highest point. At this point, the acceleration from gravity is working to change the motion of the ball from positive (upward) to negative (downward). This change is represented by the x-axis on a velocity versus time graph. As the ball changes direction, its velocity crosses the x-axis, momentarily becoming zero.

The woman needs to swim $\displaystyle 1.5km$, or $\displaystyle 1500m$. She normally swims at $\displaystyle 1.4\frac{m}{s}$, but has a $\displaystyle 0.3\frac{m}{s}$ current opposing her. The effect will make her relative velocity equal to her normal velocity, minus the current against her.

$\displaystyle 1.4\frac{m}s{}-0.3\frac{m}{s}=1.1\frac{m}{s}$

We know that velocity is a change in distance divided by a change in time. Now that we have her relative velocity and the distance traveled, we can isolate the time variable and solve for her time.

$\displaystyle v=\frac{d}{t}$

$\displaystyle t=\frac{d}{v}$

$\displaystyle t=\frac{1500m}{1.1\frac{m}{s}}$

$\displaystyle t=1364s$

To solve this problem we can simply examine each biker separately and see how long it would take them to reach the destination. Let's begin with the recreational biker. The recreational biker needs to ride $\displaystyle 3.22*10^4m$ at $\displaystyle 8.05\frac{m}{s}$. Using the definition of velocity, we can find his final time.

$\displaystyle v=\frac{d}{t}$

$\displaystyle t=\frac{d}{v}$

$\displaystyle t=\frac{3.22*10^4m}{8.05\frac{m}{s}}$

$\displaystyle t=4000s$

The professional has a distance of $\displaystyle 3.22*10^4m$, plus the $\displaystyle 1.20*10^4m$ that he's behind.

$\displaystyle 3.22*10^4m+1.20*10^4m=4.42*10^4m$

We know that the velocity of the professional biker is $\displaystyle 11.1\frac{m}{s}$. Using this velocity and his total distance, we can find the time that it takes him to reach the end of the path.

$\displaystyle v=\frac{d}{t}$

$\displaystyle t=\frac{d}{v}$

$\displaystyle t=\frac{4.42*10^4m}{11.1\frac{m}{s}}$

$\displaystyle t=3982s$

The time of the professional biker is less than that of the recreational biker, meaning that the professional will finish first. Now we need to find the distance between the two bikers at this point. Use the recreational biker's velocity and the time difference between the two bikers to solve for the distance that the recreational biker has left on the path.

$\displaystyle 4000s-3982s=18s$

The recreational biker will ride for $\displaystyle 18s$ at $\displaystyle 8.05\frac{m}{s}$ to finish the path.

$\displaystyle v=\frac{d}{t}$

$\displaystyle d=vt$

$\displaystyle d=(8.05\frac{m}{s})(18s)$

$\displaystyle d=144.9m\approx145m$

The problem gives us the distance, the acceleration due to gravity, and implies that the initial velocity of the picture is zero, as it starts at rest.

We can use the appropriate motion equation to solve for the final velocity:

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

We can use our values to solve for the time. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

$\displaystyle -1.5m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$\displaystyle -1.5m=-4.9\frac{m}{s^2}t^2$

$\displaystyle \frac{-1.5m}{-4.9\frac{m}{s^2}}=t^2$

$\displaystyle 0.306s^2=t^2$

$\displaystyle \sqrt{0.306s^2}=\sqrt{t^2}$

$\displaystyle 0.55s=t$

The ball will travel upwards to the highest point, then change direction and travel downwards. Remember that the velocity in at the highest point is equal to zero. We can use the following equation and our known data to solve for the initial velocity.

$\displaystyle V_{f}=V_{i}+at$

$\displaystyle 0\frac{m}{s}=V_{i}+(-9.8\frac{m}{s^2})(3s)$

$\displaystyle 0\frac{m}{s}=V_{i}-29.4\frac{m}{s}$

$\displaystyle 29.4\frac{m}{s}=V_{i}$

The relationship between constant velocity, distance, and time can best be illustrated as $\displaystyle \Delta x =v*t$.

We are given the velocity and the distance, allowing us to solve for the time that the ball was in motion.

$\displaystyle \Delta x =v*t$

$\displaystyle 10m =0.8\frac{m}{s}*t$

Isolate the variable for time.

$\displaystyle \frac{10m}{0.8\frac{m}{s}} =t$

$\displaystyle 12.5s=t$