The height in feet above the ground of a ball thrown upwards from the top of a building is given by s = -16x^2 + 160x + 200, where t is the time in seconds

a) what is the maximum height (in feet) of the ball?
b) what is v^-(32)

Question

The height in feet above the ground of a ball thrown upwards from the top of a building is given by s = -16x^2 + 160x + 200, where t is the time in seconds

a) what is the maximum height (in feet) of the ball?
b) what is v^-(32)

agentnao · Answer

Use a graphing calculator to solve this. It'll be the maximum point of the graph.

anonymous · Answer

use vertex formula

-b/2a

anonymous · Answer

160/2(32)

160/64

agentnao · Answer

Should be 600

agentnao · Answer

http://www.wolframalpha.com/input/?i=+-16x%5E2+%2B+160x+%2B+200

anonymous · Answer

use vertex for which problem A or B

phi · Answer

Your equation uses x, but I assume it should be t.  
so t= -b/2a,  and you must plug this value into the equation to find the maximum height.

anonymous · Answer

what about B

phi · Answer

what is v^(-32)?

anonymous · Answer

yes

agentnao · Answer

-b/2a doesn't really work for this number. The height of the top of a building won't be -160/32...

anonymous · Answer

-b/2a

160/64=2.5
plug in 2.5 into your equation

anonymous · Answer

so you take derivative of equation first?

phi · Answer

is v velocity?  then yes, take the derivative.

anonymous · Answer

no i mean for maximum height of ball

phi · Answer

t= -b/2a = 5
s = -16*5^2 + 160*5 + 200= 600

phi · Answer

Or take the derivate, set it = 0, and solve for t.  You get t=5

anonymous · Answer

What about B?????