trajectory of a particle in a vertical plane is : y=ax - bx^2. what is maximum height reached by the particle?

Question

trajectory of a particle in a vertical plane is : y=ax - bx^2. what is maximum height reached by the particle?

Answer 1

Write

y = a x - b x^2



To find the maxima minima, find the first derivative and equate that to zero.

dy/dx= a - 2bx=o

x= a/2b

The second derivate

d^2y/dx^2 = -2b

Therefore =1/2b is the maximum height. Put this value of x in the first equation. You get



y = a a/2b = a^2/2b - b a^2/4b^2 =a^2/4b

Answer 2

-b/(2a)= x value of vertex. Sub that in to get y value of y=-ba/2 +(b^3)/(4a^2)

Answer 3

the answer is a^2/4b. how to get it?