Respuesta :

irspow
This is a freefall problem with an acceleration of -10m/s (a poor approximation of g :P)

a=g

v=gt+v0

h=gt^2/2+v0t+h0  where v0 is initial velocity and h0 is initial height.

You have:

h=-5t^2+10t+3

Since h0=3, the springboard is 3 meters above the water.

...

The diver hits the water when h=0.

5t^2-10t-3=0

Using the quadratic equation for simplicity:

t=(10±√160)/10

t=(10±4√10)/10 seconds, since t>0

t≈2.26 seconds  (to nearest hundredth of a second)

So the diver hits the water about 2.26 seconds after (s)he leaves the springboard.

Otras preguntas