Question #a7a70

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Answer 1 · 2016-09-25T20:12:18

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

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

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

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

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

By kinematics equation

$v^{2} = u^{2} + 2 g h$ $v^2=u^2+2gh$

$\Rightarrow v^{2} = 0^{2} + 2 \times 3.7 \times 239$ $=>v^2=0^2+2xx3.7xx239$

$\Rightarrow v \approx 42 m/s$ $=>v~~42"m/s"$

$h = u t + \frac{1}{2} \times g \times t^{2}$ $h=ut+1/2xxgxxt^2$

$\Rightarrow 239 = 0 \times t + \frac{1}{2} \times 3.7 \times t^{2}$ $=>239=0xxt+1/2xx3.7xxt^2$

$\Rightarrow t = \sqrt{\frac{239 \times 2}{3.7}} \approx 11.4 s$ $=>t=sqrt((239xx2)/3.7)~~11.4s$