An aeroplane is moving with a velocity $u$. It drops a packet from a height $h$. The time $t$ taken by the packet in reaching the ground will be
An aeroplane is moving with a velocity $u$. It drops a packet from a height $h$. The time $t$ taken by the packet in reaching the ground will be
$\sqrt {\left( {\frac{{2g}}{h}} \right)} $
$\sqrt {\left( {\frac{{2u}}{g}} \right)} $
$\sqrt {\left( {\frac{h}{{2g}}} \right)} $
$\sqrt {\left( {\frac{{2h}}{g}} \right)} $
An aeroplane is moving with a velocity $u$. It drops a packet from a height $h$. The time $t$ taken by the packet in reaching the ground will be
The initial velocity of aeroplane is horizontal, then the vertical component of velocity of packet will be zero.
So $t = \sqrt {\frac{{2h}}{g}} $
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