Question

A particle is fired from ground with the speed u making an angle theta with the horizontal. The speed of the particle just before hitting the ground will be

(1)u
(2)u costheta
(3)u/2
(4)u sintheta​

Answer 1

Answer:

a)u

Explanation:

Given:

Initial speed of the particle =u

Angle made by particle with horizontal=$\theta$

Now when the particle falls back to the ground its displacement is 0

The initial velocity of particle in y direction=$u\sin\theta$

The initial velocity of particle in x direction=$u\cos\theta$

From the Equation of motion under gravity in y direction we have

$y=u\sin\theta-\dfrac{gt^2}{2}\\0=u\sin\theta-\dfrac{gt^2}{2}\\\\t=\dfrac{2u\sin\theta}{g}$

The y- component of velocity just before the particle hits the ground be $v_y$ given by

$v_y=u\sin\theta-g\times\dfrac{2u\sin\theta}{g}\\=-u\sin\theta$

The x-component of velocity is constant as their is no acceleration in x direction

The net velocity be $v_{net}$

$v_{net}=\sqrt{(u\cos\theta)^2 +(-u\sin\theta)^2}\\=u$

Answer 2

Answer:

THE ANSWER TO THIS QUESTION IS OPTION (1) u

Explanation:

GIVEN : The initial velocity of the particle is u and the angle it made with the horizontal is θ

The horizontal component of velocity is vₓ = u cosθ

The vertical component of velocity is vₙ = u sinθ

As in horizontal direction there is no force aₓ=0

Using the equation of motion in the horizontal direction

v=u+at

as the aₓ=0

vₓ = u cosθ

Using the equation of motion in the vertical direction

As in vertical direction aₙ = -g as g is reducing the speed in vertical direction

vₙ = u sinθ - gt

And time of flight is t=$\frac{2usinθ}{g}$

substituting

vₙ=-u sinθ

Now the velocity vector v=vₓ i+ vₙ j

Therefore v= u cosθ i + u sinθ j

Any vector of the form x i+ y j has magnitude $\sqrt{x^{2} +y^{2} }$

Using this fact v=$\sqrt{u^{2} }$

v=u

Hence I chose option (1) u