Question

Consider a situation in which a uniform gravitation field is present below line AB and
no gravity above line AB. Then the speed with which a rod shown in the figure will
strikes earth surface when released from the position shown is:
a)✓gl
b)✓2/gl
c)✓3/gl
d)✓4/gl​​

Answer 1

Answer:

Well, if we add just one line of code and grab the magnitude of that vector before normalizing it, then we'll have the distance

Answer 2

The speed at which the rod strikes the earth surface when released from the position is  3$\sqrt{gl}$

Given:

Gravitation field is present below line AB, no gravity above line AB.

To Find:

The speed with which a rod shown in the figure will strikes earth surface when released from the position

Solution:

To get the value of the initial position when it strikes the earth surface

is given by,

a = $\frac{\frac{m}{l} yg}{m}$

= $\frac{gy}{l}$

The limit tends from 0 to u,

We get,

∫ 0 to u (vdv) = $\frac{g}{l}$∫ 0 to u (ydy)

u = $\sqrt{gl}$......(1)

$v^2=u^2+2g(4l)$

u = 3$\sqrt{gl}$

Hence, The speed at which the rod strikes the earth surface when released from the position is  3$\sqrt{gl}$

#SPJ2