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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 $30\,<\rm>$

Recall that the projectiles is actually a certain form of 100 % free-slip motion having a production direction off $\theta=90$ having its own formulas .

Solution: (a) Let the base of your very well be the foundation

(a) What lengths is the ball out of the really? (b) Brand new brick in advance of returning on better, just how many moments are beyond your better?

Earliest, we discover simply how much length golf ball goes up. Recall the high section is the perfect place $v_f=0$ therefore we has\begin

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 the provider end up being the organizing part. With the help of our known beliefs, you will discover the original speed 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 https://www.datingranking.net/plano-dating 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 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: Between the supply (skin peak) together with highest area ($v=0$) pertain committed-separate kinematic equation below to obtain the most readily useful level $H$ where the golf ball reaches.\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?




















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