Question

A body is projected with a velocity of 120 ms at an angle of 30° with the horizontal. Its velocity
after 6 sec is (g = 10ms-?)​

Answer 1

AnswEr :

• Velocity of the particle (u) = 120 m/s

• Angle of Projection (∅) = 30

• Time Period = 6 seconds.

To finD :

Velocity after 6sec.

Velocity along the horizontal remains constant throughout the motion of the particle,since velocity is varying across the normal.

Initial Velocity along the Normal :

$\sf \: u.sin(30) \\ \\ \longrightarrow \: \sf \: 120 \times \dfrac{1}{2} \\ \\ \longrightarrow \sf \: 60 \: {ms}^{ - 1}$

We know that,

$\sf v_y = u_y + at$

Here,

• The body is projected upwards,thus a = - g m/s².

Now,

$\dashrightarrow \sf v_y = 60 - (10)(6) \\ \\ \dashrightarrow \boxed{ \boxed{\sf v_y = 0 \: {ms}^{ - 1} }}$

• The velocity of the particle along y-axis after 6 seconds is 0 m/s.

As stated earlier, velocity along x-axis is constant.

$\sf v_x = u_x = u.cos(30) \\ \\ \longrightarrow \sf v_x = 120 \times \dfrac{\sqrt{3}}{2} \\ \\ \longrightarrow \boxed{\boxed{\sf v_x = 60 \sqrt{3} ms^{-1} }}$

• The velocity of the particle along x-axis is 60√3 m/s.

Thus,

$\sf v_{net} = \sqrt{{v_x}^{2} + {v_y}^{2} } \\ \\ \implies \sf v_{net}= \sqrt{{v_x}^2} \\ \\ \implies v_{net} = v_x \\ \\ \implies \boxed{\boxed{\sf v_{net} = 60 \sqrt{3} ms^{-1}}}$

The net velocity after 6 seconds is 60√3 m/s

Answer 2

Explanation:

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