We have two athletes, Bryan and Matt. Bryan weighs 95 kg and Matt weighs 76 kg. Both complete a standing vertical jump. Bryan jumped 0.33 m off the floor and Matt jumped 0.45 m off the floor. Calculate power for each jump.
Most Helpful Opinions
I'd love to calculate this, however it has been years since I've solved this type of physics problem. I'd have to spend a few minutes refreshing my memory.5
its a bank holiday im not fcking doing thisfuck you matti dont like matt0
What Girls & Guys Said
I'm not doing your physics homework for you2
The only opinion from guys was selected the Most Helpful Opinion, but you can still contribute by sharing an opinion!
Click "Show More" for your mentions
Most Helpful Opinion(mho) Rate.
Are you 100% sure that you need to calculate the POWER and not the WORK (or ENERGY)?
Energy in the jump is easy to compute.
Power is not because that involves knowing time because power is rate of energy per unit time.
We can compute the time of the jump and that sort of could be used as the time, but that seems a bit complicated for a simple physics question like this.
Apparently my brother's sheet said power on it, which confused me as well cause energy definitely makes more sense, which is why I asked this question.
The energy of the jump is MgH
where M is the mass
H is the height that the guy jumps to
g = acceleration of gravity = 9.80665 m/s^2.
Now, you COULD compute the flight time.
H=v (0)*t - g*t^2/2
But, we don't know v (0).
However, dropped from H, the speed at hitting the floor is v^2=2gH
So, that's probably the speed we need.
The weight of each guy is the force, I believe, that he needs to overcome.
So, perhaps the power, P = F*V = Mg*SQRT (2gH)
So, for Bryan:
M = 95 kg
H = 0.33 m
g = 9.8 m/s^2
P = (95)(9.8)*sqrt (2*9.8*0.33) = 2367.7 W
So, for Matt:
M = 76 kg
H = 0.45 m
g = 9.8 m/s^2
P = (76)(9.8)*sqrt (2*9.8*0.45) = 6994.8 W
Yeah, go with those numbers. I am more confident in this calculation now.
Are you sure about this?
At my age, I am sure of nothing.
But, that's my answer and that is what I would put on a test (using a g of 9.8 and not 9.80665).
9.8 or 9.81 I guess.
The "official" value for the mean acceleration of gravity at the surface of the Earth is called "Standard Gravity" and is 9.80665 m/s^2.
I was talking about rounding off the numbers though. I already knew that.
standard acceleration of gravity
Numerical value 9.806 65 m s-2
Standard uncertainty (exact)
Relative standard uncertainty (exact)
Concise form 9.806 65 m s-2
Thanks a tonne!
Please let me know if I was correct.
It's been a long time...
No worries. I'll suss it out first.
So, what was the answer?