Question

A body is projected at an angle of 60° with horizontal the value of y according to the given figure

Answer 1

a body is projected at an angle 60° with horizontal.

given, horizontal range = 6m + 3m = 9m

let initial speed of body is u

then, horizontal range = u²sin2(60°)/g

or, 9m = u² sin120°/10

or, 90 = u² × sin(180 - 60°)

or, 90 = u² sin60°

or, 90 = u² × √3/2

or, u² = 60√3 ......(1)

now, applying formula, $x=u_xt+\frac{1}{2}a_xt^2$ along horizontal direction.

here, x = 6m, $u_x=ucos60^{\circ}$ and ax = 0

so, 6 = ucos60° t

or, t = 6/ucos60° = 12/u.......(2)

applying formula, $y=u_yt+\frac{1}{2}a_yt^2$

here, uy = usin60° , t = 12/u [ from equation (2) ] and ay = -g

then, y = usin60° × 12/u - 1/2 × g × 144/u²

= 12sin60° - g × 72/u²

putting g = 10m/s² and u² = 60√3 from equation (1),

y = 12 × √3/2 - 10 × 72/60√3

= 6√3 - 12/√3

= 6√3 - 4√3

= 2√3

hence, option (2) is correct choice.