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 $->h=35-1/2(9.81)t^2$

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

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

$t^2=70/9.81$

$=>t=sqrt(70/9.81) ~~2.67" seconds"$
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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 $h=35-4.9t^2$

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 + 35h = -4.9² + t + 35. How long until he reaches the water?