A particle is fired from ground with the speed u making an angle theta

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

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»

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 $v_{net}$

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