Question

a body is projected with a velocity 60m/s at 30° to horizontal. its velocity after 3s​

Answer 1

Explanation:

time=3s u=60sin30=30 a=-g so v=u+at v=30-30=0

Answer 2

Solution

Velocity after 3s would be zero

Given

• Initial Velocity,u = 60 m/s

• Angle of Projection∅ = 30°

To finD

Velocity after three seconds of projection

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The velocity vector would have two components,one along the X axis and another along the Y axis

• X axis : u cos∅

• Y axis : u sin∅

• The horizontal velocity vector remains unchanged thus the acceleration is zero

• The vertical velocity vector tends to change with respect to height

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Now,

Along X axis

$\tt \: \: u_x = u \cos( \alpha ) \\ \\ \longrightarrow \tt \: u_x = 60 \times \cos(30) \\ \\ \longrightarrow \: \boxed{ \boxed{ \tt \: u_x = 30 \sqrt{3} \: \: {ms}^{ - 1} }}$

Along Y axis

$\tt \: u_y = u \sin( \alpha ) \\ \\ \longrightarrow \: \tt \: u_y = 60 \times \sin(30) \\ \\ \longrightarrow \: \boxed{ \boxed{ \tt \: u_y = 30 \: {ms}^{ - 1} }}$

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Using the Relation,

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

Substituting the values,we get :

$\leadsto \tt \: v \: = (u \: \sin( \alpha ) ) - (10)(3) \\ \\ \leadsto \: \sf \: v = 30 - 10(3) \\ \\ \huge{ \leadsto \: \boxed{ \boxed{ \tt \: v = 0 \: {ms}^{ - 1}}}}$

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