How can I calculate how high an arrow goes when all I know is its initial speed? I'm not familiar with English physics terms so bear with me. If I shot an arrow straight up and it went off with a speed of 21 m/s, how high up would it go? (air resistance is insignificant). My textbook does not give the mass of the arrow so I don't even know how to begin solving this. All it tells me is that the initial speed is 21 m/s.
 A: At the instant you release the arrow, it begins to lose velocity due to gravitational acceleration, which can be represented by a vector pointing opposite to the arrow's direction of flight.
The arrow loses about 9.8 m/sec of its velocity every second (this is the magnitude of gravitational acceleration at the Earth's surface).  Solve for the time it would take to reach terminal vertical velocity of 0 m/sec, and calculate how far upward it would have travelled during that time.  You need to account for the constant diminution of the arrow's vertical velocity.
There is a kinematic equation derived from calculus that solves for displacement when initial velocity, time and constant acceleration are known, but I'll let you locate that.  Don't forget that acceleration in this case is negative.
A: There are some very useful elementary equations that describe basic motion with constant acceleration, and these are: $$v=u+at,$$ $$v^2=u^2+2as,$$ $$v=ut+\frac{1}{2}at^2,$$ where $u$ is initial speed, $v$ is final speed, $s$ is displacement (how far the object has moved) and $a$ is acceleration. 
You must now think about your problem to determine which quantities you know in order to determine which equation to use.


*

*You have the initial speed, $u=21m/s$.

*The ball is in the air, so gravity is acting on it. This means that the acceleration of the ball is the acceleration due to gravity (which acts down and is $a=g=9.8m/s^2$).

*If you think about when you throw a ball in the air, it slows down to a halt, and then begins falling back down. So when it is at the maximum height the final velocity is $v=0m/s$.

*You now need to calculate $s$, which is the height so we call $s=h$.
So you have $u$,$v$,$a$ and you need $s$, which means that you want to use the equation $$v^2=u^2+2as=u^2-2gh.$$ (Remember I said that $g$ is acting downwards since gravity acts downward, which is why it is negative.) Rearranging for your desired quantity you get $$h=\frac{u^2-v^2}{2g}.$$ Substituting values you get $$h=\frac{21^2-0^2}{2\times9.8}=22.5m.$$ Hope this helps, feel free to ask a question if you need me to clarify anything.
A: How long does it take to stop? $v/a$ or roughly 2 seconds. How far does it go up in 2 seconds? ${1/2}at^2$ or 20m, roughly. You figure it out exactly.
A: Besides the well written answer from Olly, I'd like to add the following :  
You can think of the problem in terms of conservation of energy.  When at the surface the arrow has zero gravitational potential energy.
Launching it with some velocity provides it with some energy which turns into Kinetic energy due to its motion.
When you reach the highest point your same kinetic energy would be converted to potential energy.  That is because at highest point you have zero velocity hence zero kinetic energy but the energy that you provided it initially has to be converted to something. In this case potential energy.  
Now you can use the standard formulas for KE and PE as follows 
At highest point.  
KE =PE 
$\dfrac{1} {2} mv^2 = mgh$
or 
$v^2/2= gh$  
Substituting v and g you get
$h=22.5 mt$
