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 is a particular types of free-fall action having a production angle from $\theta=90$ having its own formulas .
(a) What lengths ’s the baseball from the well? (b) The brick in advance of returning into better, exactly how many moments try beyond your better?
Solution: (a) Allow base of the well be the origin. Remember that highest section is the perfect place $v_f=0$ therefore we possess\begin
v_f^<2>-v_0^<2>=-2g\Delta y\\0-(30)^<2>=-2(10)(\Delta y)\\=45\,< \rm>\end
Basic, we discover how much cash length the ball goes up
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
\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 $t=3\,< \rm>$. Read More
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