Question 202853
If a juggler can toss a ball into the air with a velocity of 64ft/sec from a height of 6 ft, then what is the maximum height reached by the ball?
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{{{v[f] = v[0] - 32t}}}

where {{{v[f]}}} = the final velocity
where {{{v[0]}}} = the initial velocity 

(that's the speed at which the ball was thrown 
upward at the beginning, which is 64 ft/sec)

The ball has to stop momentarily when it gets to its maximum height.

{{{t}}} = the number of seconds it takes the ball to 
change its speed from {{{v[0]=64}}}ft/sec to {{{v[f]=0}}}  

Therefore {{{v[f]}}} = 0

Substituting into

{{{v[f] = v[0] - 32t}}}

{{{0 = 64 - 32t}}}

{{{32t = 64}}}

{{{t=64/32}}}

{{{t=2}}}.

So it reaches its maximum height in 2 seconds.

Now we use another formula to find out how high the
ball was exactly 2 seconds after it was thrown up:

{{{h=h[0]+v[0]t - 16t^2}}}

{{{h}}} = the height of the ball at time t

{{{h[0]}}} = the initial height 

(that's the height of ball at beginning of the toss, {{{h[0]=6}}}feet

and where {{{v[0]}}} = the initial velocity = {{{64}}}ft/sec

(that's the height of ball at beginning of toss)

Substitute {{{h[0]=6}}}, {{{v[0]=64}}}, {{{t=2}}}

in:

{{{h=h[0]+v[0]t - 16t^2}}}

{{{h=6+64(2) - 16(2)^2}}}

{{{h=6+128-16(4)}}}

{{{h=134-64}}}

{{{h=70}}}

So the maximum height of the ball is 70 feet

Edwin</pre>