Question 771131
h, in feet of an object above the ground is given by h(t)=-16t^2 + 64t + 190 where t is the time in seconds.  Find the time it takes the object to strike the ground and find the maximum height of the object.
I have no idea what to do with this equation.  I have tried putting h in terms of t and solving from there, but it equals zero.  I do not know what else to do.  Thanks so much.

h(t)=-16t^2 + 64t + 190 

at two times the height of the object = 0 when it starts and when it reaches the ground since this a parabola
plug h=0
-16t^2 + 64t + 190 =0

Find the roots of the equation by quadratic formula							
							
a=	-16	b=	64	c=	190		
							
b^2-4ac=	4096	-	-12160				
b^2-4ac=	16256			{{{sqrt(	16256	)}}}=	127.5
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}							
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}							
x1=(	-64	+	127.5	)/	-32		
x1=	-1.98						
x2=(	-64	-127.5	) /	-32			
x2=	5.98						
Ignore negative value			mph				
x	=	5.98 

time taken = 5.98 seconds

The maximum must occur at the vertex. So let’s find the vertex. We use the formula t =-b/2a (-64)/2*-16= 2 So we have t = 2 seconds.

So at 2 seconds the object reaches its maximum height.  we must find the value of the vertex, Find the value of h when t = 2. 

We plug this in to the equation -16*(2)^2 + 64*2 + 190 = = 254 feet. 
So the maximum height is 254 feet.