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A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to v0. Due to the wind the balloon gathers the horizontal velocity component vx = ay, where a is a constant and y is the height of ascent. Find how the following quantities depend on the height of ascent: (a) the horizontal drift of the balloon x(y); (b) the total, tangential, and normal accelerations of the balloon.

Radhika Batra , 11 Years ago
Grade 11
anser 1 Answers
Arun Bhadoria

Last Activity: 7 Years ago

(a)   When balloon is at sarting point
Vy=v°, Vx=0
since vertical velocity is constant thus vertical acceleration is 0
Now after time t 
Vy=v°, Vx =ay
Where y =v°t (vertical distance = Vert.velocity*time)
Thus Vx=av°t
         dx/dt=av°t
dx=av°t dt
Integrating both sides
X = av°t2/2
X=ay2/2v° = horizontal drift         (since t=y/v°)
(b) Now total accr. Is due to Vx
Vx =ay
dVx/dt= a(dy/dt)
Ax= a Vy =         av°        = total accr. (Since Vy =v°)
(c) total velocity= ay i + v° j  (in vector form)
Tangential accr. = Accr. Along total velocity
av° * unit vector along total velocity (dot product)
av° i . [(ay i + v°j)/ {(ay)2+(v°)2}1\2 ] .                                   (Since     accr.  dot   total velocity in vector form divided by magnitude of total velocity)
After solving you get= a2v°y/(v°2 + a2y2)½  = Tang. accr.
 
(d)  norm. accr. ={ ( Total accr.)2 -  (Tang. accr.)2}½
{(av°)2 - (Tang. accr.)2}½
do it yourself and get   
           av°2/{v°2 + a2y2}½.       Thank you...
 
 
 
 
 
 
 
 
 
 
 
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