A baseball is thrown up into the air. The ball's height in feet is modeled by the equation h(t) = -16t2 + 32t + 10. (h = height, t = time in seconds) Find the maximum height of the ball. a) 10 feet b) 2.3 feet c) 23 feet d) 26 feet
Do you know how to take derivative?
differentiate h' = -32t + 32 = 0 t = 1 plug 1 into your formula for h(t) and you have it
use formula when x=(-b/2a), y will have a max/min. in this quadratic equation, it has the max. x=(-b/2a)=-32/(2*-16)=1 and then, substitude x=1 into the equation. u can get max y=26
did u get it?
I know this is basically what Andrew said, but I am just trying to give an intuition on how it works. Since we know the ball will spend them amount of time going up as it does going down we know that the halfway point between the zeros of the function will be the maximum. So if we solve for the root and then add the roots together and divide by 2 we get the halfway point which is 1. Roots --> ((1/4)(4-(26)^(1/2))+(1/4)(4+(26)^(1/2))) = 2 and 2/2 is 1. H must reach a max at t=1 so plug this back into the formula like all the others said and you get your answer. Just a different way of thinking about problem.
the same amount of time**
@Andrewnj ya i get it thanks
@mathdood thanks a lot :)
26 feet.. any doubt ?
so u guys think the answer is 26 for sure ???? :)
do u understand?
not really :)
k.if t=1 then -16(1)^2 + 32(1) + 10=26 if t is equal to more then 1, there will be lesser amount then 26. eg: -16(2)^2 + 32(2) + 10=10 eg: -16(3)^2 + 32(3) + 10=-38 so time should be less, bt time cant be a minus figure becouse of dat i took t=1. did u get it?
It would be easiest if you understand differentiation. Do you? If not finding the midpoint of the roots is simpler but more tedious.
didnt get it..
suppose u hav y=x^n then its differentiation dy/dx will be nx^(n-1) differentiation of a number is zero. and if u hav y=nx its differentiation dy/dx will be n. is t clear upto this?
@this r u there?
yes @ ajprincess
@ILM.Riyas finally i got it thanks a lot:D
thanks ya'all r the best :)
yw:)
@ this always welcome
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