Problem (55): From the bottom of a $25\,<\rm>$ well, a stone is thrown vertically upward with an initial velocity $30\,<\rm>$
Remember your projectiles try a specific sort of free-slip motion with a release position out-of $\theta=90$ featuring its own algorithms .
(a) How long ‘s the golf ball from the better? (b) The new brick ahead of coming back with the better, exactly how many mere seconds was away from really?
Solution: (a) Allow base of your well be the foundation. Keep in mind your large part is the place $v_f=0$ so we has actually\begin
v_f^<2>-v_0^<2>=-2g\Delta y\\0-(30)^<2>=-2(10)(\Delta y)\\=45\,<\rm>\end
Very first, we discover just how much distance golf ball increases
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 https://datingranking.net/nl/alt-overzicht/ find the total time which the stone is out of the well\begin
\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: Let the origin be the throwing point. The tower’s height is $20-<\rm>$ and total time which the ball is in the air is $4\,<\rm>$. With these known values, one can find the initial 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, 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 your resource (facial skin top) together with highest area ($v=0$) apply enough time-separate kinematic picture below to discover the greatest height $H$ the spot where the baseball are at.\initiate
v^<2>-v_0^<2>=-2\,g\,\Delta y\\0-(30)^<2>=-2(10)H\\\Rightarrow H=45\,<\rm>\end
Practice Problem (59): A rock is thrown vertically upward from a height of $60\,<\rm>$ with an initial speed of $20\,<\rm>$
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.
