Question

The equation of trajectory of a projectile is given by as y = 2x- x^2 /2 .the maximum height of the projectile is ( symbols have usual meanings and si unit) a =4m. b =1m. c =2m. d = 10m

Answer 1

Answer:

Option (c) 2m

Explanation:

Given the equation of the projectile

$y=2x-\frac{x^2}{2}$

Differentiating the above w.r.t. x

We get

$\frac{dy}{dx} = 2-2x$

From applications of differentiation, we know that

For maxima or minima

$2-x = 0\\\implies x=2$

at x = 1

$\frac{d^2y}{dx^2}$ = -ive

therefore at x = 2, there is a point of maxima for the given curve

Therefore for x = 2. the maximum height

y = 2 × 2 - 2²/2

  = 4-2

  = 2 m

Answer 2

Answer:

refer to the attachment...i hope it will help