The height of an arrow shot into the air is nodeled by the equation y=64t-16t^2, where y=feet above the ground and t seconds after release. How highdoes the arrow go?

Question

anonymous · Answer

32 metres

anonymous · Answer

Show work?

anonymous · Answer

on differertitating u will get the ans

anonymous · Answer

I am only in algebra I

anonymous · Answer

@dldinatale wat can i do thn..? u want me  to do it by algebra...?

anonymous · Answer

Yes. We. eare on quadratic equations in 9th grade

anonymous · Answer

take 16 common frm right side 
nw u will get
16[4t-t^2]=y

anonymous · Answer

Just take the derivative. The derivative is just the rate of change of the equation when the instantaneous rate of change is zero the ball has stopped and will start falling down again. Just do nx^n-1 so here 64t^(1-1)-16*2t^(2-1) = 64-32t and then solve for when the equation equals zero and isolate t. This will give you the time when the ball is at it's max height then sub that time t back into the first equation which gave you the height at each moment and you will have the height right before the arrow starts to fall back down which is the max height. I got 64 but I might be doing something wrong does this look right to you nasavivek?