a ball is thrown upward. what is the intial vertical speed? the accleration of gravity is 9.8m/s^2 and max height is 6.5m. neglect air resistance
I think I'd just use v2 = u2 + 2as where v = final velocity (= 0), a = gravitational acceleration (-9.8m/s2), s = distance (= 6.5m)
0 = u2 - 2*9.8*6.5
$${\mathtt{u}} = {\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{6.5}}}} \Rightarrow {\mathtt{u}} = {\mathtt{11.287\: \!160\: \!847\: \!617\: \!969\: \!5}}$$
initial velocity ≈ 11.3 m/s
Hi there!
Let me help you out here.
There are two formula's you need here.
At the start, the ball only has kinetic energy (speed).
This kinetic energy can be calculated as
$$E_k = \frac{1}{2}mv^2$$
where Ek is the kinetic energy, m is the mass and v is the speed.
At 6.5 meters all the kinetic energy has transformed into potential energy (since we neglect air resistance) which has the following formula;
$$E_p = mgh$$
Where Ep is the potential energy, m is the mass, g is acceleration of gravity and h the height.
Therefore if we equate $$E_p = E_k$$
We have
$$\frac{1}{2}mv^2 = mgh$$
Which we van rewrite to
$$\begin{array}{lcl}
\frac{1}{2}v^2 = gh \mbox{ (divide both sides by m)}\\
v^2 = 2gh\\
v = \sqrt{2gh}\\
\mbox{ fill in g = 9.8 and h = 6.5}\\
v = \sqrt{2*9.8*6.5} \approx 11m/s
\end{array}$$
So the speed at the start was approximately 11 meters per second.
Reinout
Ok I might do it by calculus
$$\\\ddot y=-9.8\\
\dot y=-9.8t+v\\
y=-4.9t^2+vt\\
\mbox{max height when }\doty=0\qquad(y=6.5m)\\
-9.8t+v=0\\
v=9.8t\\\\
y=-4.9t^2+vt\\
6.5=-4.9t^2+9.8t^2\\
4.9t^2-6.5=0\\
t=\sqrt{\frac{6.5}{4.9}}\\\\
\mbox{Initial Velocity}=9.8\times\sqrt{\frac{6.5}{4.9}}=11.287\;\;m/s\\\\$$
I think I'd just use v2 = u2 + 2as where v = final velocity (= 0), a = gravitational acceleration (-9.8m/s2), s = distance (= 6.5m)
0 = u2 - 2*9.8*6.5
$${\mathtt{u}} = {\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{6.5}}}} \Rightarrow {\mathtt{u}} = {\mathtt{11.287\: \!160\: \!847\: \!617\: \!969\: \!5}}$$
initial velocity ≈ 11.3 m/s
I'm not sure whether this is an English expression, but in Dutch we have the expression;
'there are several ways that lead to Rome'.
I think that it pretty much sums up these answers.
I didn't know the equation Alan used.
Melody's approach is one I hadn't directly thought of, but I think I like it better than sticking your nose into physics formulas
In NSW, Australia schools.
If you are doing physics you have to use the physics formulas.
If you are doing mathematics you have to use calculus.
I think that our expression is "All roads lead to Rome".
In English the phrase is "All roads lead to Rome".
There is another (rather unpleasant) phrase: "There is more than one way to skin a cat"
The kinematics equation I used is derived directly from calculus - see my last post at http://web2.0calc.com/questions/two-students-on-a-balcony-18-8m-above-the-street-one-student-throws-a-ball-vertically-downward-13-1-m-s-at-the-same-instant-the-onther-st_1#r6
Ai, that sounds awefully unpleasant.
Please don't skin the cat!
Several of our expressions sound very strange when you literally translate them such as;
it is a truth as a cow!
or
I can do this with two finger up my nose
or
That's as correct as a bus
So given 'all craziness on a stick' are there more english expressions that are funny or strange?
I think the "skin a cat" expression probably has more to do with the "cat-o-nine-tails" (a whip used to keep recalcitrant sailors in order in days of yore) than cute furry animals!
I must be mad, but I really like the expression "I can do this with two fingers up my nose". From now on I shall use it whenever I can!!