How High Does a Jumper Raise Their Center of Gravity?

[bionut]

Homework Statement

A jumper takes off with a velocity of 5 ms at an angle of 20 degrees to the horizontal.
How high does he raise his center of gravity?

Homework Equations

The Attempt at a Solution

I know that Vh= 5scos20 = 1.7 m/s and Vv=5Csin20=4.70 m/s

using 0=Vi^2 + 2ad
0=5sin20 + 2 X-9.81d
19.62d=1.7
d=1.7/19.62
d=0.087m, but its incorrect. Am I misssing something? The only Thing I can think of is beacuse its his centre of gravity that I should treat it as a vertical launnched projectile?

 

[PeterO]

Homework Statement

A jumper takes off with a velocity of 5 ms at an angle of 20 degrees to the horizontal.
How high does he raise his center of gravity?

Homework Equations

The Attempt at a Solution

I know that Vh= 5sin20 = 1.7 m/s and Vv=5Cos20=4.70 m/s

using 0=Vi^2 + 2ad
0=5sin20 + 2 X-9.81d
19.62d=1.7
d=1.7/19.62
d=0.087m, but its incorrect. Am I misssing something? The only Thing I can think of is beacuse its his centre of gravity that I should treat it as a vertical launnched projectile?

Nothing wrong with your calculation. Since they are asking how high, could this be the high-jump and the angle is actually 20 degrees to the vertical?

 

[bionut]

thanks for your help, that's what i was thinking would it then mean that the Vv is 5Cos20 instead of 5Sin20? Is that what you are suggesting?, also the answer is 0.14m...

 

[bionut]

Can anyone else help... I still can't get 0.14m?

:-(

 

[rwisz]

I would like to clarify a few things about the given scenario and the calculations provided. Firstly, the given information does not specify the mass of the jumper, which is an important factor in determining the height of their center of gravity. Additionally, it is not clear if the 5 m/s velocity given is the initial velocity or the final velocity.

Assuming that the given velocity is the initial velocity and the mass of the jumper is 70 kg (average weight of an adult male), the height of the center of gravity can be calculated using the equation d = v^2sin^2θ/2g, where d is the height, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity (9.81 m/s^2).

Plugging in the values, we get d = (5 m/s)^2sin^2(20°)/(2*9.81 m/s^2) = 0.61 m. This means that the jumper raises their center of gravity by 0.61 meters. However, this calculation assumes that the jumper has a uniform distribution of mass and that their center of gravity is located at their geometric center.

If the given velocity is the final velocity and the jumper is able to maintain this velocity throughout the jump, the height of the center of gravity can be calculated using the equation d = vf^2sin^2θ/2g, where vf is the final velocity. In this case, the height would be higher than 0.61 m, as the jumper would have covered more distance with a higher final velocity.

In conclusion, as a scientist, I would recommend providing more specific information and clarifying any assumptions made in order to accurately determine the height of the jumper's center of gravity.