Question

A particle is projected with velocity v and v vector is equals to 3 icap - 2 j cap + 2 k cap metre per second and the constant acceleration except acting on the particle is a vector is equals to minus 6 icap + 2 j cap - 4 k cap metre per second then the path of a projectile is what

Answer 1

Answer:

Parabolic Cylinder

Explanation:

1) We have,

$\vec{v}=3i-2j+2k\\ \\ \vec{a}=-6i+2j-4k$

2) Using Newtons Equation,

$x=v_xt+\frac{1}{2}a_xt^2\\ \\=3t-\frac{1}{2}*6*t^2\\ \\=3t-3t^2$

Also,

$y=v_yt+\frac{1}{2}a_yt^2\\ \\=-2t+\frac{1}{2}*2*t^2\\ \\=-2t+t^2$

Again,

$z=v_zt+\frac{1}{2}a_zt^2\\ \\=2t+\frac{1}{2}*-4*t^2\\ \\=2t-2t^2$

3) Getting t from 'x' and 'y' ,we get

$t=-(\frac{x}{3}+y)$

4) Substituting above value of t in 'z', we get

$z=-2(\frac{x}{3}+y)-2(\frac{x}{3}+y)^2\\ \\= -\frac{2x}{3}-2y-2y^2-\frac{2x^2}{9}-\frac{4xy}{3}\\ \\ =>z+2y^2+\frac{2x^2}{9}+\frac{4xy}{3}+\frac{2x}{3}=0$

Hence, Required path of projectile is parabolic cylinder.

Answer 2

Answer:it will be a straight line