Question 1005463
{{{h=-(1/2)gt^2+v[0]t+h[0]}}} is a general formula.
 
For the first problem we will set:
{{{t}}}= time from the moment the penny is dropped (we will measure it in seconds, because it will be a short time),
{{{h[0]=1250}}}= height (in feet) at the moment, {{{t=0}}} , when the penny is dropped,
{{{v[0]=0}}}= the initial velocity (in feet/second) of the penny at {{{t=0}}} (since it was "dropped", it was not thrown up or down with any initial velocity).
We will use the acceleration of gravity in {{{ft/second^2}}} , to agree with the other units:
{{{g=32}}} = acceleration of gravity in {{{ft/second^2}}} .
Substituting, we get
{{{h=-(1/2)32t^2+0*t+1250}}}--->{{{h=-16t^2+1250}}}
 
1a. For {{{t=3}}} , {{{h=-16*3^2+1250=-16*9+1250=-144+1250=1106}}} ,
so the height of the penny after three seconds is {{{highlight(1106feet)}}} .
1b. When the penny hits the ground, {{{h=0}}} . Then,
{{{0=-16t^2+1250}}}--->{{{16t^2=1250}}}--->{{{t^2=1250/16}}}--->{{{t^2=78.125}}}--->{{{t=sqrt(78.125)}}}--->{{{t=about 8.84}}} ,
so the penny hits the ground after about {{{highlight(8.85seconds)}}} .
 
 
For the second problem we will set:
{{{t}}}= time from the moment the ball is tossed (we will measure it in seconds, because it will be a short time),
{{{v[0]=10}}}= the initial velocity (in meters/second) of the ball at {{{t=0}}} (since it was "dropped", it was not thrown up or down with any initial velocity).


{{{red(IF)}}} (as I originally misread) {{{h[0]=495}}}= height (in meters) at the moment, {{{t=0}}} , when the ball is tossed,
We will use the acceleration of gravity in {{{meters/second^2}}} , to agree with the other units:
{{{g=9.8}}} = acceleration of gravity in {{{meters/second^2}}} .
Substituting, we get
{{{h=-(1/2)9.8t^2+10t+495}}}--->{{{h=-4.9t^2+10t+495}}}
2. For {{{t=3}}} ,
{{{h=-4.9*3^2+10*3+495}}}-->{{{h=-4.9*9+30+495}}}-->{{{h=-44.1+30+495}}}-->{{{h=480.9}}} ,
so the height of the ball after three seconds is {{{highlight(480.9meters)}}} .
2b. When the ball hits the ground, {{{h=0}}} . Then,
{{{0=-4.9t^2+10t+495}}} .
Solving the quadratic equation we get two solutions, but the negative solution does not make sense, and needs to be discarded.
So, {{{t=(-10 + sqrt(10^2-4*(-4.9)*495 ))/(2*4.9)}}}
{{{t=(-10 + sqrt(100+9702))/9.8}}}
{{{t=(-10 + sqrt(9802))/9.8}}}
{{{t=about(-10 + 99)/9.8=89/9.8=about9.08}}} ,
so the ball hits the ground after about {{{highlight(9.08seconds)}}} .


{{{red(IF)}}} it was really meant as  {{{h[0]=495}}}= height (in feet),
we have a units mismatch.
We have to get everything in feet, or everything in meters.
Since I am a supporter of the SI system of units, I will convert {{{495 feet}}} to meters.
{{{495 feet=(495 feet)(0.3048meters/"1 foot")=150.9m}}}(rounded), so
{{{h[0]=150.9}}}= height (in meters) at the moment, {{{t=0}}} , when the ball is tossed,
We will use the acceleration of gravity in {{{meters/second^2}}} , to agree with the other units:
{{{g=9.8}}} = acceleration of gravity in {{{meters/second^2}}} .
Substituting, we get
{{{h=-(1/2)9.8t^2+10t+150.9}}}--->{{{h=-4.9t^2+10t+150.9}}}
2. For {{{t=3}}} ,
{{{h=-4.9*3^2+10*3+150.9}}}-->{{{h=-4.9*9+30+150.9}}}-->{{{h=-44.1+30+495}}}-->{{{h=136.8}}} ,
so the height of the ball after three seconds is {{{highlight(136.8meters)}}} .
2b. When the ball hits the ground, {{{h=0}}} . Then,
{{{0=-4.9t^2+10t+150.9}}} .
Solving the quadratic equation we get two solutions, but the negative solution does not make sense, and needs to be discarded.
So, {{{t=(-10 + sqrt(10^2-4*(-4.9)*150.9 ))/(2*4.9)}}}
{{{t=(-10 + sqrt(100+2957.6))/9.8}}}
{{{t=(-10 + sqrt(3057.6))/9.8}}}
{{{t=about(-10 + 55.30)/9.8=45.30/9.8=about4.62}}} ,
so the ball hits the ground after about {{{highlight(4.62seconds)}}} .