“m dv/dt=-mg-kv v(0)=0” (Wherek=k_1)
This can be re arranged to dv/dt=-g-Dv/m, if we make –g the subject dv/dt+kv/m=-g
If we let only dv/dt and v be left in the equation and swap out all the other terms we can use an integrating factor. dv/dt+vL(t)=N(t) Where L(t)=k/m and N(t)=-g
This means v=1/u ∫▒〖N(t)u dt〗
Where u=e^∫▒L(t)dx =e^(k/m) dt =e^(k/m t)
This in turn means v=e^(-k/m t) ∫▒〖(-g)e^(k/m t) dt〗 or e^(k/m t) v=∫▒〖(-g)e^(k/m t) dt〗 e^(k/m t) v=(-gm)/k e^(k/m t)+C
To work out C we can use the initial condition v(0)=0 so t(0)=0 e^(k/m(0)) 0=(-gm)/k e^(k/m(0))+C
This means C=gm/k
Returning to the Full equation again e^(k/m t) v=(-gm)/k e^(k/m t)+gm/k v(t)=(-gm)/k+gm/k e^(--k/m t) v(t)=gm/k(e^(- k/m t)-1) For 0≤t
(as we are measuring velocity downwards to be negative when t becomes a great number the e^(-k/m t) gets smaller making the number negative.)
The position can be found by integrating the velocity, with the same conditions x(0)=0
P(t)=∫▒v(t) dt Or P(t)=∫▒〖(-gm)/k+gm/k e^(-k/m t) 〗 dt
P(t)=(-gm)/k t-(gm^2)/k^2 e^(-k/m t)+C
Again if we let P(0)=0 we are able to find C
P(0)=(-gm)/k (0)-(gm^2)/k^2 e^(-k/m(0))+C
P(0)=-(gm^2)/k^2 +C we can say P(0)=P_0
This means C=P_0+(gm^2)/k^2
Finally our equation for position becomes
P(t)=(-gm)/k t-(gm^2)/k^2 e^(-k/m t)+P_0+(gm^2)/k^2
“P(t)=P_0-gm/k t-(gm^2)/k^2 〖(e〗^(-k/m t)-1) 0≤t
Modelling the Deployed chute
As we had before we had the condition v(0)=0 we can now change that to v〖(t〗_0)=v_0 (as t0 is the time the parachute is deployed by the parachutist) v〖(t〗_0)=v_0=-mg/k-gAe^(-k/m t_0 )
Making A the subject A=-e^(-k/m t_0 ) (m/k+v_0/g)
If we Put A back into the or...
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The small pilot chute is affected by an extra drag force attempting to keep it stationary. When this force and the force of the falling diver create enough tension in the line connected to the pilot chute, the deployment bag containing the main canopy is unstowed.
This question refers to the example data given below. Using the rate law and the experimental values given below, calculate k.
Firstly we shall look upon the formula that shows how acceleration is 'supposed' to occur.
This paper will explain a few of the key concepts behind the physics of skydiving. First we will explore why a skydiver accelerates after he leaps out of the plane before his jump, second we will try and explain the drag forces effecting the skydiver, and lastly we will attempt to explain how terminal velocity works.
Galileo demonstrated that an object falling only under the influence of gravity will experience a constant acceleration, i.e.., it gains the same amount of velocity for every additional second that it falls. (5)
Everest, F. and Pohlmann, K. 2009. Master Handbook of Acoustics. 5th ed. of the book. Johnlsayers.com.
Barbara Mowat and Paul Warstine. New York: Washington Press, 1992. Slethaug, Gordon. A. See "Lecture Notes" for ENGL1007.
When t = 6.08, is negative means at this time, height is maximum = 60.08
(2003, 4). Hans Peter's Mathematical, Technical, Historical and Linguistic Omnium Gatherum. PMM physics. About friction. Retrieved November 15, 2013, from http://www.hp-gramatke.net/pmm_physics/english/page0150.htm
The period of the simple pendulum oscillations increases as the length of the pendulum increases. The acceleration due to gravity can be found experimentally from the dependent period and independent length.
Olenick, P. Richard, The Mechanical Universe: Introduction to Mechanics and Heat, Cambridge: Cambridge University Press 1985
As for the magnitude of µt or μ should be in the range of 0 and 1 (Lin and Kao, 2014), in which the value of 0 means that the real output at time t is precisely equal to the real output of the previous period, t-1. While if 1 indicates that the real output is equivalent to the desired output. Conversely, µt is a function of St, a vector of the variable, affecting the speed of adjustment of a firm, and γ is the unknown parameters. Therefore, in order to return to the conception of the PAV theory, the Equation (II.3) can be rewritten as
Projectile motion is used in our daily lives, from war, to the path of the water in the water fountain, to sports. When using a water fountain or hose, projectile motion can be used to describe the path and motion of the water. This technology was created by finding the angle at which the water would come out at a maximum height and the person using it would be able to drink it without leaning over too much. These types of projectile motion will be further explored and analyzed in this assessment.
In conclusion, we learned that in order to solve any angular projectile motion problem, one must find the vertical and horizontal components of the velocity of that object by using simple trigonometry. Then one must calculate time that the object was in the air for, insert it into the displacement equation in order to eliminate one of the unknowns. Finally, we can solve for how far the object travelled.