Problem (55): From the bottom of a $25\,$ well, a stone is thrown vertically upward with an initial velocity $30\,$

Problem (55): From the bottom of a $25\,<\rm>$ well, a stone is thrown vertically upward with an initial velocity $30\,<\rm>$

Keep in mind your projectiles are a particular kind of free-slide actions which have a launch position from $\theta=90$ using its own algorithms .

Solution: (a) Allow bottom of one’s very well be the foundation

(a) What lengths ‘s the basketball out of the well? (b) The newest brick ahead of returning to your better, exactly how many seconds is actually beyond your well?

First, we find exactly how much range golf ball increases. Bear in mind that highest section is where $v_f=0$ therefore we has actually\start

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 provider end up being the organizing part. With the help of our understood philosophy, you will discover the first velocity as \begin

\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, https://www.sugardaddydates.org/sugar-daddies-canada/halifax/ 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 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: Within provider (surface height) together with highest section ($v=0$) implement the full time-independent kinematic picture below to find the most useful peak $H$ in which the basketball has reached.\start

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?