Question

A particle is projected at an angle of elevation sin`[ 4÷5]and it's range on the horizontal plane is 3km.The initial velocity of projecton is

Answer 1

Hello Dear,

◆ Answer -

u = 177 m/s

◆ Explaination -

# Given -

θ = sin⁻¹(4/5)

R = 3 km = 3000 m

# Solution -

Given that -

θ = sin⁻¹(4/5)

sinθ = 4/5

So now,

cosθ = √(1-sin²θ)

cosθ = √(1-16/25)

cosθ = 3/5

Horizontal range of projectile is -

R = u².sin(2θ)/g

R = u².sin(2θ)/g

u² = R.g / (2sinθ.cosθ)

u² = 3000 × 10 / (2 × 4/5 × 3/5)

u² = 30000×25 / 24

u² = 31250

u = √31250

u = 176.78 m/s

Thus, initial velocity of projection is 177 m/s.

Thanks dear...