requirement number 3 troubling me, kind of confused should I get the second derivative will do it or dividing max height of the bullet over the max horizontal distance I got from the past 2 steps? the problem as follows: a bullet is shot from a submarine which is stationary on the sea level. the path of the bullet above the sea level can be modelled by the function V=-x^2+100x-1600 where V m represents the height of the bullet above the sea level, and the x m represents the horizontal distance of the bullet from an observer.
(a) Find x when V attains its maximum
(b) Hence, show that the maximum height of the bullet above the sea level is 900 m
(c) Find the horizontal distance of the bullet travelled above the sea level
i solved as follow:
(a) the first derivative of the equation = 0
the first derivative = -2x+100
-2x+100 = 0
-2x = -100
x = 50 m
(b) Vmax = -(50)^2+100(50)-1600 = 900 m
(c) i'm kinda puzzeled should i get the second derivative which will leave me with -2 and it will be the answer
or
divide Vmax over x of the max height = 900/50 = 18
or
solve the original equation V=-x^2+100x-1600 where V is 0 so x will be either 20 or 80 and we accept 20 m as it's smaller than X max and the solution will be x=20 ?
