show that time of ascent is equal to the time of descent
Answers
Answer:
:)
Explanation:
Let us take air resistance proportional to velocity which is often the case when velocity is not large. The equation of motion for upward motion is acceleration a= dv/dt=-g-kv……………(1). k is constant of proportionality . Therefore,
dv/(g+kv)=-dt. Integrating from v=u to v=0 and t=0 to t=t1, we have
t1 =(1/k)[ln(1+ku/g)]………..(1)
For downward motion, we have to know the height from where the body after time t1 returns back.Now, from eq.(1),
dv/(g+kv)=-t, we have t=(1/k)[ln(1+kv/g) or v=g/k[exp(kt)-1]. Then ,dy/dt=g/k[exp(kt)-1]. This can be integrated from t=0 to t=t1 and height can be found. After this, we write the equation for downward motion: dv/dt=g-kv and obtain v as function of time. In this function we write v=dy/dt and again integrate from y=height obtained above to y =0 and find time t2 of descent . I have tried here to give you hint of the solution and have shown nature of problem. I can not write complete lengthy calculations.
Let us take simple example of constant resistance producing constant retardation a.
For upward motion, resultant retardation will be (g+a). Then time of ascent will be
t1=u/(g+a)……………(1). The height acquired is h=u^2/2(g+a)………….(2)
The time t . of descent is obtained from h=(1/2)(g-a) t^2……………..(3).
From eqs.(2) and (3), we have u^2/(g+a)=(g-a)t^2………………(4)
But from eq.(1). u=(g+a)t1. Substituting this value of u in eq.(4) , we get
(g+a)^2t1^2/(g+a)=(g-a)t^2. OR
(g+a)t1^2=(g-a)t^^2. This gives, t1/t=sqrt[(g-a)/(g+a)]<1
Or t1<t. The time of ascent is less than the time of descent.
Answer:
ta=td
Explanation:
time of ascent = time of descent
hence, showed