Can anyone help? (1 Viewer)

[747captain]

Q: In Acapulco, on the coast of Mexico, professional high divers plunge from a height of 36metres above the water. Estimate:

1. The length of the time interal during which the divers fall through the air
2. the speed with which the divers enter the water.

Assume that throughout their dive, the divers are falling vertically from rest with an acceleration of 9.8m/s^-2.

If anyone can explain to me how you work it out, I would appreciate it.

 

[watatank]

Q: In Acapulco, on the coast of Mexico, professional high divers plunge from a height of 36metres above the water. Estimate:

1. The length of the time interal during which the divers fall through the air
2. the speed with which the divers enter the water.

Assume that throughout their dive, the divers are falling vertically from rest with an acceleration of 9.8/s^-2.

If anyone can explain to me how you work it out, I would appreciate it.

a = 9.8, s = 36, u = 0

v^2 = u^2 + 2as (u = 0)

v = sqrt(2*9.8*36)

= 26.56

v = u + at

t = (v- u)/a

=26.56/9.8

=2.71 sec

 

[747captain]

thanx but i don't get it,

but where does the "s" come from. the "s" represents time doesn't it? i think u forgot that 9.8 is actually 9.8m/s^-2.

a= 9.8m/s^-2
u= 0
v = ?
s? = (time?)

Why do we need to use: v^2 = u^2 + 2as (u = 0) ????????????????????

 

[helper]

s is displacement not time.

u=0 as it is assuming they are just falling, not diving down.

"falling vertically from rest"

The squared for a is in the units, you don't apply it to the formula

 

[747captain]

oh ok.

I think i got it. See below steps:

1. we use the formula v^2 = u^2 + 2as because we are trying to find the speed of the divers.
2. we already know that u=0, a=9.8, and s=36m(displacement)
3. v^2 = (0^2) + (2*9.8*36)
4. v = sqrt705.6
5. v approx = 26.56m/s^-1

then using the fact that v = 26.56m/s:

1. v = u + at
2. 26.56 = 0 + (9.8*t)
3. 26.56 - 0 = 9.8*t
4. 26.56/9.8 = t
5. t = 2.71sec

Thankyou so much helper and watatank

:wave:cheers