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A ball of mass M is thrown vertically upwards. Another ball of mass2M is thrown at an angle $θ$ to vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio:

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1:2

2:1

1:1

1:cos $θ$

answer is C.

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Detailed Solution

$H_{\max} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{g}{g} {(\frac{2 usinθ}{g})}^{2} = \frac{{gT}^{2}}{g}$

$H_{1} = \frac{{u_{1}}^{2}}{2 g}$

$H_{2} = \frac{{u_{2}}^{2} \sin^{2} α}{2 g} [w h e r e, α = 90^{0} - θ]$

$T_{1} = \frac{2 u_{1}}{g}; T_{2} = \frac{2 u_{2} \sin α}{g}$

$\frac{2 u_{1}}{g} = \frac{2 u_{2} \sin α}{g} \Rightarrow \frac{u_{1}}{u_{2}} = \sin α$

$\frac{H_{1}}{H_{2}} = \frac{{u_{1}}^{2}}{{u_{2}}^{2} \sin^{2} α} = 1$

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