To solve this problem, remember that when something is thrown vertically, its velocity will be $\displaystyle 0$ at the highest point. Using this principle, we know the initial velocity, final velocity (zero), and the acceleration. With the appropriate motion equation, we can solve for the distance traveled.

The best equation for this problem is:

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

Use the given values to solve for the distance.

$\displaystyle 0\frac{m}{s}^2=3.3\frac{m}{s}+2(-9.8\frac{m}{s^2})\Delta y$

$\displaystyle -3.3\frac{m}{s}=(-19.6\frac{m}{s^2})\Delta y$

$\displaystyle \frac{-3.3\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$\displaystyle 0.17m=\Delta y$

The mass of an object is completely unrelated to its free-fall motion. The equation for the vertical motion for an object in freefall is:

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

Notice, there is no mention of mass anywhere in this equation. The only thing that affects the time an object takes to hit the ground is the acceleration due to gravity and the distance travelled. Since these objects travel the same distance and are affected by the same gravitational force, they will fall for the same amount of time and hit the ground together.

The equation for the vertical motion for an object in freefall is:

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

Notice that there is no place for mass anywhere in this equation. This means that the two balls will be in the air for the same amount of time. We simply need to use the distance and acceleration to solve for the time.

Remember that even though the height is $\displaystyle 30m$, the DISPLACEMENT will be $\displaystyle -30m$. Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

$\displaystyle -30m=\frac{1}{2}(-9.8\frac{m}{s^2 })t^2$

$\displaystyle \frac{-30m}{-9.8\frac{m}{s^2}}=\frac{1}{2}t^2$

$\displaystyle 3.06s^2=\frac{1}{2}t^2$

$\displaystyle 2*3.06s^2=t^2$

$\displaystyle 6.12s^2=t^2$

$\displaystyle \sqrt{6.12s^2}=t^2$

$\displaystyle 2.47s=t$

The equation for the vertical motion for an object in freefall is:

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

Notice that there is no place for mass anywhere in this equation. This means that the two balls will be in the air for the same amount of time. We simply need to use the distance and acceleration to solve for the time.

Remember that even though the height is $\displaystyle 30m$, the DISPLACEMENT will be $\displaystyle -30m$. Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

$\displaystyle -30m=\frac{1}{2}(-9.8\frac{m}{s^2 })t^2$

$\displaystyle \frac{-30m}{-9.8\frac{m}{s^2}}=\frac{1}{2}t^2$

$\displaystyle 3.06s^2=\frac{1}{2}t^2$

$\displaystyle 2*3.06s^2=t^2$

$\displaystyle 6.12s^2=t^2$

$\displaystyle \sqrt{6.12s^2}=t^2$

$\displaystyle 2.47s=t$

When dealing with simple vertical motion, the final velocity can be found from the initial velocity, acceleration, and time.

$\displaystyle v_f=v_i+at$

Remember that because the ball is dropped from rest, the initial velocity will be zero.

Notice that mass is not a variable in this calculation; the final velocity will be the same for any mass dropped from rest that is in the air for $\displaystyle 18s$.

Use the values given in the question to solve for the final velocity.

$\displaystyle v_f=v_i+at$

$\displaystyle v_f=0\frac{m}{s}+(-9.8\frac{m}{s^2})(18s)$

$\displaystyle v_f=-176.4\frac{m}{s}$

Note that the velocity is negative because the object is traveling downward.

When dealing with simple vertical motion, the final velocity can be found from the initial velocity, acceleration, and time.

$\displaystyle v_f=v_i+at$

Remember that because the ball is dropped from rest, the initial velocity will be zero.

Notice that mass is not a variable in this calculation; the final velocity will be the same for any mass dropped from rest that is in the air for $\displaystyle 18s$.

Use the values given in the question to solve for the final velocity.

$\displaystyle v_f=v_i+at$

$\displaystyle v_f=0\frac{m}{s}+(-9.8\frac{m}{s^2})(18s)$

$\displaystyle v_f=-176.4\frac{m}{s}$

Note that the velocity is negative because the object is traveling downward.

Acceleration is the change in velocity per unit time.

$\displaystyle a=\frac{v_2-v_1}{t}$

Since the velocity does not change from one moment to the next, then there must be no net acceleration on the object.

$\displaystyle a=\frac{3\frac{m}{s}-3\frac{m}{s}}{t}=\frac{0\frac{m}{s}}{t}$

$\displaystyle a=0\frac{m}{s^2}$

For this problem, we use the formula $\displaystyle v_f=v_i+at$.

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

$\displaystyle v_f=9\frac{m}{s}+(0.5\frac{m}{s^2}*1.5s)$

$\displaystyle v_f=9\frac{m}{s}+0.75\frac{m}{s}$

$\displaystyle v_f=9.75\frac{m}{s}$

The relationship between distance, velocity, and time is given by the formula for velocity:

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

We can rearrange this equation to solve for the distance.

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

Using the values given for the velocity and the time, we can find the distance traveled.

$\displaystyle \Delta x=0.11\frac{m}{s}*3s$

$\displaystyle \Delta x =0.33m$

The relationship between distance, velocity, and time is given by the equation for velocity:

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

We can rearrange this formula for solve for the time.

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

Using the given values for the velocity and distance traveled, we can solve for the time.

$\displaystyle \Delta t =\frac{12m}{2.22\frac{m}{s}}$

$\displaystyle \Delta t=5.41s$