Question

Two Stones are thrown vertically upwards simultaneously with their velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u²1:u²2. Assume upward acceleration is -g and downward acceleration is +g.

Answer 1

ANSWER
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$for \: \: 1st \: stone \: - \\ \\ u \: = u1 \: \\ a \: = \: - g \\ v = 0 \: (velocity \: at \: the \: highest \: point \: ) \\ s = \: h1 \\ \\ {v}^{2} - {u}^{2} = \: 2as \: \: we \: get \\ 0 \: - { {u}^{2} = \: - 2gh}^{2} \\ h1 = \: \frac{u {1}^{2} }{2g} \\ \\ for \: 2nd \: stone - \\ \\ u \: = \: u2 \\ a \: = \: - g \\ v = 0 \\ s = h2 \\ \\ {v}^{2} - {u}^{2} = \: 2as \: \: we \: get \\ 0 - {u}^{2} = \: - 2gh \\ h2 \: = \frac{ {u2}^{2} }{2g}$

Then , h1 : h2

=
$\frac{ {u1}^{2} }{2g} is \: to \: \frac{ {u2}^{2} }{2g} \\ \\ = > \: \frac{ {u1}^{2} }{2g } \times \: \frac{2g}{ {u2}^{2} }$

=> u1² : u2² ( HENCE PROVED )

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