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!!