Question

The equation of trajectory of a projectile is given by y = `"x"/sqrt3 - "gx"^2/20`, where x and y are in metres. The maximum range of the projectile is:

विकल्प

•  `8/3 "m"`

• `4/3 "m"`

• `3/4 "m"`

• `3/8 "m"`

Answer 1

`4/3 "m"`

Explanation:

Comparing the given equation with the equation of trajectory of a projectile,

y = `"x" tan θ - "gx"^2/(2"u"^2 cos^2θ)`

we get, tan θ = `1/sqrt3` ⇒ θ = 30°

and 2u² cos² θ = 20

⇒ u² = `20/(2 cos^2θ)`

= `40/3`

Now, R_max = `"u"^2/"g"`

= `40/(3 xx 10)`

= `4/3 "m"`