A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function
h = −16t2 + 36t + 5. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?
This can be solved in two ways: With heavy tools or with just algebra. What is your level? Have you studied calculus?
With pure algebra: We need to find the maximum of the function <span>h = −16t^2 + 36t + 5 Lets take out -1 for simplicity: </span><span>h = −(16t2 - 36t - 5) For now lets just work with this: </span>16t^2 - 36t - 5 16t^2=(4t)^2 (4t-x)^2= 16t^2-2*4xt+x^2 we have -36t so x should be 4.5 as 2*4*4.5=36
Lets see what we have now: 16t^2 - 36t - 5= (4t-4.5)^2 is this true? No but close (4t-4.5)^2= 16t^2- 2*4*4.5t +4.5^2= 16t^2-36t+20.25
16t^2 - 36t - 5 and 16t^2-36t+20.25 nearl the same just take away 25.25 from the right hand side
Getting long, just stay with me: 16t^2 - 36t - 5= (4t-4.5)^2 - 25.25
h= -{(4t-4.5)^2 -25.25} h=-(4t-4.5)^2 + 25.25
We want to find the maximum of this function. -(4t-4.5)^2 this bit is always negative or 0, so it maximum is when it is 0. Solve: 4t-4.5=0 t=1,125
