Question

projectile is projected vertically upward The time of ascent of the projectile is 2s. The time of descent of the projectile is​

Answer 1

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 2

Answer:

2seconds

Explanation:

The time taken by body to reach maximum height is called time of ascent.

                                      $t= \frac{V_{0} sin \alpha }{g}$

The time taken by the body to reach from maximum height to the lowest level of trajectory is called the time of descent.

                                      $t= \frac{V_{0} sin \alpha }{g}$

As we can see  the time of ascent is equal to the time of descent.

Therefore the time of descent of the projectile will be 2 seconds.

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