Physics Two Dimensional Motion Question 211
Question: A balloon starts rising from the surface of the earth. The ascension rate is constant and equal to $ {{\text{v}} _{\text{0}}} $ . Due to the wind the balloon gathered the horizontal velocity component $ {{\text{v}} _{\text{x}}}\text{= ay} $ , where a is a constant and y is the height of ascent. The tangential, acceleration of the balloon is trough but
Options:
A) $ \frac{a^{2}y}{v _{0}} $
B) $\frac{a^2 y}{\sqrt{1 + (\text{ay} + v_0)^2}}$
C) $\frac{a^2 y}{\sqrt{1 + v_0^2}}$
D) $ \frac{a^2 v_0}{\sqrt{1 + (2y + a)^2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Since velocity in vertical direction is constant,
Therefore ${\text{a}}{y} = \frac{d{v{y}}}{dt} = 0$
The acceleration in horizontal direction,
$ a _{x}=\frac{dv _{x}}{dt}=\frac{d(av _{0}t)}{dt}=av _{0} $
$ a=\sqrt{a _{x}^{2}+a _{y}^{2}}=\sqrt{{{( av _{0} )}^{2}}+0}=av _{0} $
The total acceleration is $ av _{0} $ and directed along horizontal direction.
Let $ \theta $ is the angle that the resultant velocity makes with horizontal, then
Normal acceleration $ a _{n}=a\sin \theta $ and tangential acceleration
$ a _{t}=a\cos \theta ,\text{ we have x}\text{=}\frac{ay^{2}}{2{{\text{v}} _{\text{0}}}} $ .
$ \text{or }\text{y=}\sqrt{\frac{2xv _{0}}{a}} $
Differentiating both side of equation (iii) w.r.t. x,
We get $ 1=\frac{a}{2v _{0}}\times 2y\times \frac{dy}{dx} $
$ \text{or }\frac{dy}{dx}=\frac{v _{0}}{ay}=\tan \theta $
Now $ a _{x}=a\sin \theta =av _{0}\times \frac{( v _{0}/ay )}{\sqrt{1+{{( \frac{v _{0}}{ay} )}^{2}}}} $
$ =\frac{av _{0}}{\sqrt{1+{{( \frac{ay}{v _{0}} )}^{2}}}} $
$ a _{t}=a\cos \theta =av _{0}\times \frac{1}{\sqrt{1+{{( \frac{v _{0}}{ay} )}^{2}}}}=av _{0} $
$ \frac{ay}{\sqrt{(ay)^2 + v_0^2}} = \frac{q^2 y}{\sqrt{1 + \left(\frac{ay}{v_0}\right)^2}}$
BETA
