Question #a7a70

Question

6009 views around the world

You can reuse this answer
Creative Commons License

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#