Question

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ANSWER THIS

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Answer 1

Hola!

Let the the wire is held at an angle θ relative to the ground.

Thus,
We can use physics of an inclined plane.

ATQ,

cos ϕ = $\frac{u - v}{|| u || || v ||}$

$proj_u$ = $\frac{u - v}{|| v ||}$ = cos ϕ || u ||

∵ Angle to Gravitational Force and force opposite the normal is θ. We know that, since the force opposite the normal and force angle difference is 90

Thus,
Project vector of force of gravity onto the wire direction is then cos (90−θ) = sin θ || u ||

Substituting || u || = m × g

$F_w$ = mg sin θ

Thus,
by Newton’s Second Law, acceleration is...

$a_w$ = g sin θ

Here,

$v_0$ = 5 m/s n' a = $a_w$
where,

$a_w$ is Acceleration along the wire.

[ Note : g = $10 ms^{-2}$ ]

Using Kinematics Eqn.

$v_0$ + at = $v_1$

We Get,

Velocity at r, $v_R$ = [ 5 + 10t sin θ ] m/s

Hope it helps! :D