SATHEE: Kinematics Question 661

Kinematics Question 661

Question: A balloon starts r it’sing from the surface of the earth. The ascension rate it’s constant and equal to $v_{0}$ . Due to the wind the balloon gathered the horizontal velocity component $v_{x} = ay$ , where a is a constant and y it’s the height of ascent. The tangential, acceleration of the balloon it’s trough but

Options:

A) $ {{a^{2}}y/{{v _{0}} $

B) $ {{a^{2}}y/\sqrt{1+{{( \text{ay+}{{v _{0}} )}^{2}}} $

C) $ {{a^{2}}y/\sqrt{1+{{v _{0}}^{2}} $

D) $ {{a^{2}}v _{0}/\sqrt{1+{{( \text{2y+a} )}^{2}}} $

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Answer:

Correct Answer: B

Solution:

[b] Since velocity in vertical direction is constant,

$∴ a_{y} = \frac{d v_{y}}{dt} =0$

The acceleration in horizontal direction,

$a_{x} = \frac{d v_{x}}{d t} = \frac{d (a v_{0} t)}{d t} = a v_{0}$

$a = \sqrt{a_{x}^{2} + a_{y}^{2}} = \sqrt{{(a v_{0})}^{2} + 0} = a v_{0}$

The total acceleration is $a v_{0}$ and directed along horizontal direction.

Let $θ$ it’s the angle that the resultant velocity makes with horizontal, then

Normal acceleration $a_{n} = a \sin θ$ and tangential acceleration

$a_{t} = a \cos θ, we have x = \frac{a y^{2}}{2 v_{0}}$ .

$or y= \sqrt{\frac{2 x v_{0}}{a}}$

Differentiating both side of equation (iii) w.r.t. x,

We get $1 = \frac{a}{2 v_{0}} \times 2 y \times \frac{d y}{d x}$

$or \frac{d y}{d x} = \frac{v_{0}}{a y} = \tan θ$

Now $a_{x} = a \sin θ = a v_{0} \times \frac{(v_{0} / a y)}{\sqrt{1 + {(\frac{v_{0}}{a y})}^{2}}}$

$= \frac{a v_{0}}{\sqrt{1 + {(\frac{a y}{v_{0}})}^{2}}}$

$a_{t} = a \cos θ = a v_{0} \times \frac{1}{\sqrt{1 + {(\frac{v_{0}}{a y})}^{2}}} = a v_{0}$

$\frac{a y}{\sqrt{{(a y)}^{2} + v_{0}^{2}}} = \frac{q^{2} y}{\sqrt{1 + {(\frac{a y}{v_{0}})}^{2}}}$

Kinematics Question 661

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Answer:

Solution:

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Kinematics Question 661

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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