Let
u and
θ be the velocity of projection and angle of projection, respechively.
Given that, horizontal component of velocity.
ux=u‌cos‌θ=3m∕s Equation of projectile motion is given by
y=12x−x2 ...(i)
We know,
General equation of projectile motion,
y=x‌tan‌θ− ...(ii)
Comparing Eqs. (i) and (ii), we get
tanθ = 12
⇒=12 sin‌θ=12‌cos‌θ Multiplying on both side by
(u) u‌sin‌θ=12(u‌cos‌θ)=12×3 i.e.
u‌sin‌θ=36m∕s Now, using the expression of range.
R= R=| 2u2‌sin‌θ‌cos‌θ |
| g |
R=| 2(u‌sin‌θ)(u‌cos‌θ) |
| g |
[Using identity
sin‌2‌θ=2‌sin‌θ‌cos‌θ ]
Substituting the values, we get
R= = 21.6 m