An Olympic diver falls from rest from the platform 10m high. At what velocity does the diver hit the water is acceleration in 9.8m/s^2.
Do I assume "rest" means a initial velocity of 0? Could the formula be:
d= (vf^2-vi^2)/2(a) ?
as in d=10, a=9.8 vi=0
vf = √(d x 2(a)+ vi^2
vf= √(10 x 2(9.8) + 0^2
ans:v=14m/s
I'm not sure if this is right, the number seems too large... please help, thank you!
d = (1/2)(9.8)t^2
10 = 4.9 t^2
t = 1.43 seconds falling
v = a t
v = 9.8 (1.43) = 14 m/s , yes right
Your way is fine. I just cheated to get it faster.
Well, first of all, I have to commend your math skills! Your calculations seem to be on point, but let's double-check just to be sure. We don't want our beloved Olympic diver to go splat with a wrong answer!
So, using the formula you provided, let's plug in the values:
d = 10m (as given)
a = 9.8m/s^2 (as given)
vi = 0m/s (since the diver is at rest)
Applying the formula, we get:
vf = √(d x 2a + vi^2)
vf = √(10m x 2(9.8m/s^2) + 0m/s)^2
vf = √(10m x 19.6m/s^2)
vf = √(196m^2/s^2)
vf = 14m/s
Huh, would you look at that! Your initial suspicion was right. The number seems too large, but it's actually correct. The diver hits the water with a velocity of 14m/s. That's quite a splash!
So, off they go, diving into glory and watery shenanigans. Keep up the good work with your mathematical acrobatics!
Yes, you are correct that "rest" means the initial velocity of the diver is 0. The formula you mentioned, d = (vf^2 - vi^2) / (2a), can indeed be used to find the final velocity (vf) of the diver.
Let's plug in the values given:
d = 10m (height of the platform)
a = 9.8m/s^2 (acceleration due to gravity)
vi = 0m/s (initial velocity)
Using the formula, we have:
10 = (vf^2 - 0^2) / (2 * 9.8)
10 = vf^2 / 19.6
vf^2 = 196
vf = √196
vf = 14m/s
Therefore, the diver hits the water with a velocity of 14m/s. Your calculation is correct, and the resulting velocity is indeed 14m/s, which means the diver hits the water at this speed after falling from the 10m high platform.
Yes, you are correct in assuming that "rest" means an initial velocity of 0. When an object falls from rest, its initial velocity is 0.
To solve this problem, you can use the equation:
v_f = √(v_i^2 + 2ad)
where:
v_f is the final velocity (in this case, the velocity at which the diver hits the water),
v_i is the initial velocity (which is 0),
a is the acceleration (which is 9.8 m/s^2),
and d is the displacement (which is 10 m in this case).
Plugging in the values, you get:
v_f = √(0^2 + 2(9.8)(10))
v_f = √(0 + 196)
v_f = √196
v_f = 14 m/s
So, your calculation is correct. The diver hits the water with a velocity of 14 m/s.