Question 42073
I can solve the 1st one
 {{{s = -16t^2 + v[0]t + s[0]}}}
v(0) = 64 ft/sec
{{{s = -16t^2 + 64t }}}
v(0) = 64 ft/sec
after 1 sec
{{{s = -16(1)^2 + 64(1) }}}
{{{s = -16 + 64 }}}
{{{s = 48 }}}
The ball is 48 ft high after 1 sec
How long will it take to hit the ground?
The curve for this equation is a parabola with time
plotted horizontally and distance vertically
You want to find BOTH times when s=0. The first 
is t=0, when the ball leaves the throwers hand.
The second time s=0 is when the ball returns to
ground.
{{{s = -16t^2 + 64t }}}
{{{0 = -16t^2 + 64t }}}
rewrite
{{{64t - 16t^2 = 0}}}
{{{16t(4 - t) = 0}}}
the roots are 
t = 0 (ball leaves hand)
t = 4 (ball returns to ground)
The ball takes 4 sec to return to earth
What is the maximum height of the ball? 
What time will the maximum height be attained?
The vertex of the parabola is the max height
The vertex occurs midway between t=0 and t=4, or t=2
{{{s = -16(2)^2 + 64(2) }}}
{{{s = -16*4 + 128 }}}
{{{s = -64 + 128}}}
{{{s = 64}}}
The max height is 64 ft, 2 sec after ball is thrown