Question

With what minimum speed must a particle be projected from origin so that it is able
to pass through a given point (30m, 40m). Take g = 10 m/s2.
1) 60 m/s
2) 30 m/s
3) 50 m/s
4)40 m/s​

Answer 1

Given:

A particle is projected such that it passes through a point (30m,40m).

To find:

Minimum velocity of projection.

Calculation:

General Equation for projectile is:

$y = x \tan( \theta) - \dfrac{g {x}^{2} }{2 {u}^{2} { \cos}^{2} ( \theta) }$

Putting value of x and y:

$= > 40 = 30 \tan( \theta) - \dfrac{4500 }{ {u}^{2} { \cos}^{2} ( \theta) }$

$= > 40 = 30 \tan( \theta) - \dfrac{4500 \: { \sec}^{2}( \theta) }{ {u}^{2} }$

$= > 40 = 30 \tan( \theta) - \dfrac{4500 \: \{ 1 + { \tan}^{2}( \theta) \} }{ {u}^{2} }$

$= > 40 - 30 \tan( \theta) = - \dfrac{4500 \: \{ 1 + { \tan}^{2}( \theta) \} }{ {u}^{2} }$

$= > {u}^{2} = - \dfrac{4500 \: \{ 1 + { \tan}^{2}( \theta) \} }{ 40 - 30 \tan( \theta) }$

$= > {u}^{2} = \dfrac{4500 \: \{ 1 + { \tan}^{2}( \theta) \} }{ 30 \tan( \theta) - 40 }$

Minimum value of the trigonometric complex is:

$= > {u}^{2} = \dfrac{4500}{5}$

$= > {u}^{2} = 900$

$= > u = 30 \: m {s}^{ - 1}$

So, final answer is:

$\boxed{ \bold{ \red{ \large{ u = 30 \: m {s}^{ - 1} }}}}$

Answer 2

Answer:

30 m/s

Explanation:

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