Question

a body is projected with velocity u at an angle of 60 to the horizontal . the time interval after which it will be moving in a direction of 30 to horizontal is ​

Answer 1

Answer:

$\displaystyle t_f=u \left( 0.073\right) \ sec.$

Explanation:

Given :

Initial velocity ( u ) = u m / sec.

First case when angle is 60.

$\displaystyle t_1=\frac{2u\sin\theta}{g}\\\\ \displaystyle t_1=\frac{2u\sin60}{g}\\\\\displaystyle t_1=\frac{2u\times\dfrac{\sqrt3}{2} }{g}\\\\\displaystyle t_1=\frac{u\times\sqrt3}{g}$

Second case when angle is 30.

$\displaystyle t_2=\frac{2u\sin\theta}{g}\\\\ \displaystyle t_2=\frac{2u\sin30}{g}\\\\\displaystyle t_2=\frac{2u\times\dfrac{1}{2} }{g}\\\\\displaystyle t_2=\frac{u}{g}$

Now

$\displaystyle t_f=\frac{t_1}{t_2}$

$\displaystyle t_1-t_2=\frac{u\times\sqrt3}{g}-\displaystyle \frac{u}{g}\\\\\displaystyle t_f=\frac{u\times\sqrt3-u}{g}\\\\\displaystyle t_f=\frac{u( \sqrt3-1)}{g}\\\\\displaystyle \text{Now put g=10 and \sqrt3=1.73 }\\\\\displaystyle t_f=\frac{u(1.73-1)}{10}\\\\\displaystyle t_f=\frac{u(0.73)}{10}\\\\\displaystyle t_f=u \ 0.073 \ sec.$

Thus we get answer.

Answer 2

Explanation:

u/g√3 is the answer for your question