Question

for an object projected from ground with speed u horizontal range is two times the max. height attained.The horizontal range of object is

Answer 1

2u^2 Sinx Cosx / g = 2u^2 Sin^2x / (2g)
Tanx = 2
Sin^2 x = 5

Horizontal range of object will be = 2 * (2u^2 Sin^2x)/(2g)
= 2u^2 Sin^2x / g
= 2u^2 * 5 / g
= 10u^2 / g
= u^2 …………… [If g ≈ 10m/s^2]