KCET 2015 Physics Solved Paper

Velocity of particle = v
Horizontal range = 2 × Greatest height attained
Now,Range =‌

v2‌sin‌2‌θ

g

Height =‌

v2sin2θ

2g

Using, sin 2θ = 2 sin θ cos θ, we get
⇒tan‌θ=2
⇒sin‌θ=‌

2

√5

and cos‌θ=‌

1

√5

Therefore,
Range⇒ Horizontal range =

4

5

v2

g