Question

A particle is projected at an angle of elevation
sin
and it's range on the horizontal plane is
3 km. The initial velocity of projection is
(1) 160 m/s
(2) 175 m/s
(3) 180 m/s
(4) 230 m/s​

Answer 1

you didn't mention about angle , just given *sin* . let α is the angle of elevation or particle projected at an angle α with horizontal.

so, horizontal range of particle, R = u²sin2α/g

where u is initial velocity of particle, and g is acceleration due to gravity.

given, horizontal range = 3km = 3000m [ as we know, 1 km = 1000m]

so, 3000m = u²sin2α/g

or, 3000g/sin2α = u²

if we assume g = 10m/s²

then, 30000/sin2α = u²

hence, velocity of particle , u = $\sqrt{\frac{30000}{sin2\alpha}}$

put the value of α, you will get the answer.