AP EAPCET 24-Aug-2021 Shift 2 Paper

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−

3

4

x2 ...(i)
We know,
General equation of projectile motion,
y=x‌tan‌θ−

gx2

2u2cos2θ

...(ii)
Comparing Eqs. (i) and (ii), we get
tanθ = 12
⇒

sin‌θ

cos‌θ

=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=

u2‌sin‌2‌θ

g

R=

2u2‌sin‌θ‌cos‌θ

g

R=

2(u‌sin‌θ)(u‌cos‌θ)

g

[Using identity sin‌2‌θ=2‌sin‌θ‌cos‌θ ]
Substituting the values, we get
R=

2×36×3

10

= 21.6 m