Question

A body is projected upward with velocity 50m/s at angle 30 degree with the horizontal, so its velocity after 2 sec equal?

Answer 1

Answer:

• Velocity (v) after 2 Seconds will be 43.5 m/s.

Given:

1. Initial velocity (u) = 50 m/s.
2. Time period (t) = 2 seconds.
3. Angle of projection (θ) = 30°.

Explanation:

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This is a Case of Projectile motion.

$\boxed{\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\vector(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){ \theta }\put(0.5,1.8){u}\put(3.15,1.5){\vector(0,1){0.5}}\put(3.15,1.5){\vector(0,-1){0.5}}\put(3.3,1.4){H}\put(4.3,0.7){\vector(-1,0){4}}\put(4.3,0.7){\vector(1,0){2}}\put(3.3,0.3){R}\end{picture}}}$

We Will solve this problem with Resolving the motion into Components.

Y - Direction:-

Applying First kinematic equation for Y - Direction.

$\large\bigstar \: {\boxed{\tt v_y = u_y - gt}}$

$\bold{Here}\begin{cases}\tt{v_y} \text{ Denotes Final Velocity in Y - direction} \\ \tt{u_y} \text{ Denotes Initial velocity in Y - direction} \\ \text{g Denotes Acceleration due to gravity} \\ \text{t Denotes Time period}\end{cases}$

Now,

$\large{\boxed{\tt v_y = u_y - gt}}$

$\large{\tt \longmapsto v_y = u sin \theta - gt}$

Substituting the values,

$\large{\tt \longmapsto v_y = 50 \times sin 30 - 10 \times 2}$

• g = 10 m/s².
• t = 2 seconds.
• u = 50 m/s.

$\large{\tt \longmapsto v_y = 50 \times \dfrac{1}{2} - 10 \times 2}$

$\large{\tt \longmapsto v_y = \dfrac{50}{2} - 20}$

$\large{\tt \longmapsto v_y = \cancel{\dfrac{50}{2}} - 20}$

$\large{\tt \longmapsto v_y = 25 - 20}$

$\large\longmapsto{\underline{\boxed{\tt v_y = 5 \: m/s}}}$

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X - Direction:-

Applying First kinematic equation for X - Direction.

$\large\bigstar \: {\boxed{\tt v_x = u_x}}$

$\bold{Here}\begin{cases}\tt{v_x} \text{ Denotes Final Velocity in x - direction} \\ \tt{u_x} \text{ Denotes Initial velocity in x - direction}\end{cases}$

Now,

$\large{\boxed{\tt v_x = u_x}}$

As the Value of g (Acceleration due to gravity) Remains Constant in X- Direction (Horizontally).

$\large{\tt \longmapsto v_x = u cos \theta}$

Therefore,

$\large{\tt \longmapsto v_x = 50 \times cos 30}$

$\large{\tt \longmapsto v_x = 50 \times \dfrac{\sqrt{3}}{2}}$

$\large{\tt \longmapsto v_x = \dfrac{50 \sqrt{3}}{2}}$

$\large{\tt \longmapsto v_x = \cancel{\dfrac{50 \sqrt{3}}{2}}}$

$\large\longmapsto{\underline{\boxed{\tt v_x = 25 \sqrt{3} \: m/s}}}$

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For Final Velocity:-

$\large\bigstar \: {\boxed{\tt v = \sqrt{(v_x)^2 + (v_y)^2}}}$

Substituting the values,

$\large{\tt \longmapsto v = \sqrt{(25 \sqrt{3})^2 + (5)^2}}$

$\large{\tt \longmapsto v = \sqrt{(625 \times 3) + 25}}$

$\large{\tt \longmapsto v = \sqrt{1875 + 25}}$

$\large{\tt \longmapsto v = \sqrt{1900}}$

$\large\longmapsto{\underline{\boxed{\red{\tt v = 43.5 \: m/s}}}}$

∴ Velocity (v) after 2 Seconds will be 43.5 m/s.

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Answer 2

$\Huge{\underline{\underline{\mathfrak{Question \colon }}}}$

A body is projected upward with velocity 50m/s at angle 30 degree with the horizontal, so it's velocity after 2s of projection is :

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From the Question,

• Initial Velocity (u) = 50 m/s

• Angle of Projection (∅) = 30°

• Time Taken (t) = 2 s

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Note

• Velocity along the x - axis would be uniform throughout

• Velocity along the y - axis is subjected to change

• Velocity after 2s would be the magnitude of sum of both velocities

$\boxed{\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\vector(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){ \theta }\put(0.5,1.8){u}\put(3.15,1.5){\vector(0,1){0.5}}\put(3.15,1.5){\vector(0,-1){0.5}}\put(3.3,1.4){H}\put(4.3,0.7){\vector(-1,0){4}}\put(4.3,0.7){\vector(1,0){2}}\put(3.3,0.3){R}\end{picture}}}$

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Along Y - Axis

$\large{ \boxed{ \boxed{ \tt{ {v}_{y} = {u}_{y} + at }}}}$

• Velocity along y - axis is taken as u sin∅

• Acceleration would be "- g"

Substituting the values,we get :

$\large{ \sf{ {v}_{y} = u \: \times \sin 30 - gt }} \\ \\ \large{ \hookrightarrow \: \sf{ {v}_{y} } = 50 \times \dfrac{1}{2} - 10 \times 2 } \\ \\ \large{ \hookrightarrow \: \sf{ {v}_{y} = 25 - 20}} \\ \\ \large{ \hookrightarrow \: \underline{ \boxed{\sf{ {v}_{y} = 5 \: {ms}^{ - 1} }}}}$

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Along X - Axis

• Velocity along X - axis is taken as u cos∅

• Acceleration would be zero,as velocity is uniform

Thus,

$\large{ \boxed{ \boxed{ \tt{ {v}_{x} = u \: \cos( \theta) }}}}$

Putting the values,we get :

$\large{ \hookrightarrow \: \sf{ {v}_{x} = 50 \times \dfrac{ \sqrt{3} }{2} }} \\ \\ \large{ \hookrightarrow \: \underline{ \boxed{\sf{ {v}_{x} = 25 \sqrt{3} \: m {s}^{ - 1} }}}}$

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Now,

$\displaystyle { \large{ \sf{v = \sqrt{ {{v}_{x}}{}^{2} + {v}_{y}{}^{2} }}}} \\ \\ \displaystyle{ \large{\longrightarrow \: \sf{v = \sqrt{(25 \sqrt{3}) {}^{2} + {5}^{2} } }}} \\ \\ \displaystyle{ \large{ \longrightarrow \ \sf{v = \sqrt{1900} }}} \\ \\ \displaystyle{ \large{ \longrightarrow \ \boxed{ \boxed{ \sf{v = 43.5 \: ms {}^{ - 1} }}}}}$

The velocity of the particle after 2 seconds of projection would be 43.5 m/s

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