Question

case
[
R=frac{u^{2} sin 2 theta}{g} quad H=frac{u^{2} sin ^{2} theta}{2 g}
]
let speed of ( A=u ) speedof ( B=4 )
2
[
begin{aligned}
frac{H_{D}}{H_{B}}=frac{U^{prime} sin ^{2} theta_{A} / 2 g}{psi^{2} sin ^{2} theta_{B} / 2 g}=frac{sin ^{2} theta_{A}}{sin ^{2} theta_{B}}=frac{3}{1} & frac{sin theta_{A}}{sin theta_{B}}=frac{sqrt{3}}{1}
theta_{A}=60^{circ}
end{aligned}
]
Speed of ( A=2 mu )
[
theta_{2}=30^{circ}
]
[
R=frac{4 y^{2} sin 2 theta_{A} / g}{psi^{2} sin 2 theta_{B} / g}=frac{4 sin 120^{circ}}{sin 60^{circ}}=4: 1
]

# (c) 600 (a) 90 21. Two particles A and B are projected with same speed so that the ratio of their maximum heights reached is 3 : 1. If the speed of A is doubled without altering other parameters, the ratio of the horizontal ranges attained by A and B is (a) 1:1 (b) 2:1 (c) 4:1 (d) 3:2 (e) 4:3 atile of an angle of

Solution