Problem (55): From the bottom of a $25\,<\rm>$ well, a stone is thrown vertically upward with an initial velocity $30\,<\rm>$
Remember that projectiles is a particular particular free-fall action that have a release angle out-of $\theta=90$ along with its very own formulas .

## Solution: (a) Allow bottom of your well be the foundation

(a) How far is the baseball out from the really? (b) Brand new brick prior to going back towards really, how many mere seconds is actually beyond your really?

Basic, we discover just how much length the ball increases. Keep in mind that the high area is the place $v_f=0$ so we features\initiate

v_f^<2>-v_0^<2>=-2g\Delta y\\0-(30)^<2>=-2(10)(\Delta y)\\=45\,<\rm>\end

Of this height $25\,<\rm>$ is for well’s height so the stone is $20\,<\rm>$ outside of the well.

v_i^<2>-v_0^<2>=-2g\Delta y\\v_i^<2>-(30)^<2>=-2(10)(25)\\\Rightarrow v_i=+20\,<\rm>\end

where $v_i$ is the velocity just before leaving the well which can serve as initial velocity for the second part to find the total time which the stone is out of the well\begin

## The tower’s height is $20-<\rm>$ and total time which the ball is in the air is $4\,<\rm>$

\Delta y=-\frac 12 gt^<2>+v_0 t\\0=-\frac 12 (-10)t^<2>+20\,(2)\end

Solving for $t$, one can obtain the required time is $t=4\,<\rm>$.

Problem (56): From the top of a $20-<\rm>$ tower, a small ball is thrown vertically upward. If $4\,<\rm>$ after throwing it hit the ground, how many seconds before striking to the surface does the ball meet the initial launching point again? (Air resistance is neglected and $g=10\,<\rm>$).

Solution: Allow source end up being the organizing part. With your known philosophy, one can find the first velocity as \start

\Delta y=-\frac 12 gt^<2>+v_0\,t\\-25=-\frac 12 (10)(4)^<2>+v_0\,(4)\\\Rightarrow v_0=15\,<\rm>\end

When the ball returns to its initial point, its total displacement is zero i.e. $\Delta y=0$ so we can use the following kinematic equation to find the total time to return to the starting point \begin

\Delta y=-\frac 12 gt^<2>+v_0\,t\\0=-\frac 12\,(10)t^<2>+(15)\,t\end

Rearranging and solving for $t$, we get $t=3\,<\rm>$.

Problem (57): A rock is thrown vertically upward into the air. It reaches the height of $40\,<\rm>$ from the surface at times $t_1=2\,<\rm>$ and $t_2$. Find $t_2$ and determine the greatest height reached by the rock (neglect air resistance and let $g=10\,<\rm>$).

Solution: Let the trowing point (surface of ground) be the origin. Between origin and the point with known values $h=4\,<\rm>$, $t=2\,<\rm>$ one can write down the kinematic equation $\Delta y=-\frac 12 gt^<2>+v_0\,t$ to find the initial velocity as\begin

\Delta y=-\frac 12 gt^<2>+v_0\,t\\40=-\frac 12\,(10)(2)^<2>+v_0\,(2)\\\Rightarrow v_0=30\,<\rm>\end

Now we are going to find the times when the rock reaches the height $40\,<\rm>$ (Recall that when an object is thrown upward, it passes through every point twice). Applying the same equation above, we get \begin

\Delta y=-\frac 12 gt^<2>+v_0\,t\\40=-\frac 12\,(10)t^<2>+30\,t\end

Rearranging and solving for $t$ using quadratic formula, two times are obtained i.e. $t_1=2\,<\rm>$ and $t_2=4\,<\rm>$. The sugar baby in Iowa greatest height is where the vertical velocity becomes zero so we have \begin

v_f^<2>-v_i^<2>=2(-g)\Delta y\\0-(30)^<2>=2(-10)\Delta y\\\Rightarrow \Delta y=45\,<\rm>\end

Thus, the highest point located $H=45\,<\rm>$ above the ground.

Problem (58): A ball is launched with an initial velocity of $30\,<\rm>$ vertically upward. How long will it take to reaches $20\,<\rm>$ below the highest point for the first time? (neglect air resistance and assume $g=10\,<\rm>$).

Solution: Amongst the supply (body height) therefore the large section ($v=0$) implement the amount of time-separate kinematic picture below to get the better height $H$ where baseball has reached.\begin

v^<2>-v_0^<2>=-2\,g\,\Delta y\\0-(30)^<2>=-2(10)H\\\Rightarrow H=45\,<\rm>\end

The point $20\,<\rm>$ below $H$ has height of $h=45-20=25\,<\rm>$. The time needed for reaching that point is obtained as\begin

\Delta y=-\frac 12\,g\,t^<2>+v_0\,t\\25=-\frac 12\,(10)\,t^<2>+30\,(t)\end

Solving for $t$ (using quadratic formula), we get $t_1=1\,<\rm>$ and $t_2=5\,<\rm>$ one for up way and the second for down way.

Practice Problem (59): A rock is thrown vertically upward from a height of $60\,<\rm>$ with an initial speed of $20\,<\rm>$. Find the ratio of displacement in the third second to the displacement in the last second of the motion?