A diver dives into the sea from a height of 35m. His height h meters t seconds after leaving the cliff is given by #h = -4.9² + t + 35#. How long until he reaches the water?

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Answer 1 · 2016-07-28T21:41:39

2.67 seconds

The diver starts at the height of 35 metres. At the point of impact with the water his height is 0.

#0=-4.9^2+t+35#

#color(red)("Is this equation correct?"#

As I understand things

Acceleration will be 9.81 metres per second squared

So height will be starting distance -( mean velocity times time)

mean velocity #-> 1/2 a t-> "distance" = 1/2(9.81)t^2#

So height #->h=35-1/2(9.81)t^2#

#=>0=35-1/2(9.81)t^2#

#1/2(9.81)t^2=35#

#t^2=70/9.81#

#=>t=sqrt(70/9.81) ~~2.67" seconds"#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Standard equation is #s=ut+1/2at^2#

#u=0#

#s=1/2at^2 = 4.9t^2" " larr 1/2a = 9.8/2=4.905 " call it "4.9#

so #h=35-4.9t^2#