Question #a7a70

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Answer 1 · 2016-09-25T20:12:18

Given
$u->"Initial velocity of camera"=0$

$g->"Acceleration due to gravity"=3.7m/s^2$

$h->"Height of fall of camera"=239m$

$v->"Final velocity of camera"=?$

$t->"Time of fall of camera"=?$

By kinematics equation

$v^2=u^2+2gh$

$=>v^2=0^2+2xx3.7xx239$

$=>v~~42"m/s"$

$h=ut+1/2xxgxxt^2$

$=>239=0xxt+1/2xx3.7xxt^2$

$=>t=sqrt((239xx2)/3.7)~~11.4s$