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 craigslist personals gay Los Angeles.
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 resource (surface peak) and also the highest part ($v=0$) apply the time-separate kinematic picture lower than to obtain the finest height $H$ where baseball is located at.\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?