A balloon starts rising from the surface of the Earth. The ascertion rate is constant and equal to `v_0`. Due to the wind the balloon

A balloon starts rising from the surface of the Earth. The ascertion r

1.	A balloon starts rising from the surface of the Earth. The ascertion rate is constant and equal to `v_0`. Due to the wind the balloon
Answer» Correct Answer - (a) `x=(a//2v_0)y^2`; (b) `w=av_0`, `w_tau=a^2y//sqrt(1+(ay//v_0)^2)`, `w_n=av_0//sqrt(1+(ay//v_0)^2)`. According to the problem (a) `(dy)/(dt)=v_0` or `dy=v_0dt` Integrating `underset0oversetyintdy=v_0underset0overset tint dt` or `y=v_0t` (1) And also we have `(dx)/(dt)=ay` or `dx=aydt=av_0tdt` (using 1) So, `underset(0)overset(x)intdx=av_0underset0oversettint t dt`, or, `x=1/2av_0t^2=1/2(ay^2)/(v_0)` (using 1) (b) According to the problem `v_y=v_0` and `v_x=ay` (2) So, `v=sqrt(v_x^2+v_y^2)=sqrt(v_0^2+a^2y^2)` Therefore `w_t=(dv)/(dt)=(a^2y)/(sqrt(v_0^2+ay^2))(dy)/(dt)=(a^2y)/(sqrt(1+(ay//v_0)^2)` Diff. Eq. (2) with respect to time. `(dv_y)/(dt)=w_y=0` and `(dv_x)/(dt)=w_x=a(dy)/(dt)=av_0` So, `w=\|w_x\|=av_0` Hence `w_n=sqrt(w^2-w_t^2)=sqrt(a^2v_0^2-(a^4y^2)/(1+(ay//v_0)^2))=(av_0)/(sqrt(1+(ay//v_0)^2)`

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A balloon starts rising from the surface of the Earth. The ascertion rate is constant and equal to `v_0`. Due to the wind the balloon

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