A diver springs upward with a speed of 2m/s. The diving board is located 5 meters above the swimming pool,?

Question

Please slove:

A diver springs upward with a speed of 2m/s. The diving board is located 5 meters above the swimming pool, how long does it take for the diver to hit the water after leaving the diving board? Also how fast does she hit the water?

Answer 1

Assuming no air resistance

1. First we have to compute how high the diver can jump up
v=gt1 - > h1 =.5 gt1^2
t1=v/g then h1=.5 g (v/g)^2
or h1=(1/2g) v^2

2. H=h1+ 5m=(1/2g) v^2+5 this is the total height the diver will reach before the dive

3. And finally to answer the question we have to determine the time for the free fall using maximum height H=.5 gt^2 ant substituting that time into equation for velocity V=gt we obtain the answer

If H=.5 gt^2 then t=sqrt(2 H/g)

then V=gt= g sqrt(2 H/g)=
V=sqrt(2 H g)
V= sqrt(2 g ((5+ (1/2g) v^2)=
V=SQRT(2 x 9.81( 5+2^2/(2x9.81))=
V=10.1 m/s

Answer 2

from this formula you can solve for (t) : y=(-1/2)g*t^2 + v0t + y0
if you plug in the values you will get: 0=(-1/2)g*t^2 + 2 t + 5
and solve for t
then you can find the velocity where it hits the water (ground 0) from the following formula: v = g*t + v0 and you should plug in the t value which you got from the last step, you can even find the value where the diver reaches the highest point from the same formula. for doing so you need to set the v equal to zero.
have fun

Answer 3

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Answer 4

Wow, do you get a lot of physics homework?