If I toss a ball in the air at 32 feet per second how high will it go?

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Clarification of Question by kenpoman42-ga on 17 Feb 2006 22:39 PST

please show the math!!!!

Hello and thank you for your question.
Click here:
http://www.ncsu.edu/felder-public/kenny/images/physicist/rockthrow.gif

The formula that relates velocity, uniform acceleration, and distance (s) is
(v final)^2 = (v initial)^2 + 2 * a * s

so in your case since the ball will have v=0 at the top of its path,

 0 = 32^2 + 2 * 32 * s

or 

  64s = 32^2 
     s = 16 feet

There's a rule of symmetry that says this is the same as the height
you would drop a ball from that would give it v=32 when it hits the
ground.

If you can't assume knowing the above formula, you can get the same
answer in two steps:
Step 1:  How much time?
  v = at
  32 = 32t
   t = 1 second
Step 2:  How far?
  s = 1/2 * a * t^2
  s = 16

If you want to read more about this problem, take a look at
http://www.ncsu.edu/felder-public/kenny/papers/physicist.html

Cheers
Google Answers Researcher
Richard-ga

Well the relationship between acceleration velocity and placement is
that each is the antiderivitive of the one before it so:
Acc due to gravity = -32 ft/sec^2
integrate with respect to t and velocity is -32t + c
setting t to zero and knowing that original velocity was 32 ft/sec we
have v(0)=0+c=32 ft/sec so c=32 and v(t) = -32t + 32
to find placement we integrate again and get
-16t^2 + 32t + c
setting placement at time zero we can solve for c by
s(0)=0=0+0+c   => s(t)=-16t^2+32t
since we are concerned with placement at time 1
s(1)=-16(1)^2+32(1)= 16 Feet, so ignoring wind resistance and holding
acceleration due to gravity to -32 ft/sec^2 the ball with be 16 feet
higher from the place it left your hand and have a velocity of 0 feet
per second with acceleration of -32 feet per second squared.  I hope
this is what you had in mind by "show the math"

~mathisfun

I would be careful with the equation of s=1/2*a*t^2 in the future,
this is true for dropping a ball or transversely throwing the ball to
its pinnacle, however would not work for instance if you through the
ball in the air at the speed of 32 ft/sec and wanted to know where it
was 1.5 seconds later.  In that case calc will quickly show for given
acceleration constant "a" we have
a(t) = a
v(t) = at + c  where c is initial velocity
s(t) = 1/2*a*t^2 + Vo*t + c, where in this case c is the initial
placement.  which leads us to:
s(t) = (1/2 * a * t^2) + (Vo * t) + So 
where a is acceleratoin, Vo is original velocity, So is original
placement and t is of course time.
Example:  Say you are standing 60 feet above the ground and throw a
ball up at (+)32 feet per second with gravity pulling it down at
(-)32ft/sec^2 and whant to know where the ball is relative to the
ground after 3 seconds.
s(3) = (1/2)*(-32)*(9) + 32*3 + 60
     = 12 feet above the ground.