Question

a person throws a bottle into a dustbin at same height as he is 2m away at an angle of 45 degree. the velocity of thrown

Answer 1

Given :

The horizontal range of bottle=R=2m
R=u²sin2θ/g
2=u²sin2x45°/g
2=u²sin90°/g
2=u²/g
u=√2g

∴the velocity of thrown is √2g

Answer 2

$\displaystyle \text{It's given,}\\\text{Horizontal Range of the bottle = 2m}\\\text{It make angle of }45^o\\\\\text{We know,}\\R = \frac{u^2\sin2\theta}{g} \\2 = \frac{u^2\sin2\times45^o}{g}\\2 = \frac{u^2\sin90^o}{g}\\2 = \frac{u^2}{g} \\u = \sqrt{2g} \\\\\text{Hence the velocity of thrown} = \sqrt{2g}$