Answer to Question #236648 in Mechanics | Relativity for Arjun

Question

A particle is projected horizontally with a speed 20m/s from the top of a tower.after what time will the velocity of the particle be 45 degree angle from the initial direction of projection

Answer 1

$\vec v_x= 20\frac{m}{s}$

$g = 9.8 \frac{m}{s^2}$

$\vec v_y =\frac{gt^2}{2}$

$\vec{v} =\vec v_x+\vec v_y$

$\alpha= \angle(\vec v,\vec v_x)=45 \degree$

$\vec v_y =\tan \alpha*\vec{v_x}= v_x=20\frac{m}{s^2}$

$t = \sqrt{\frac{2v_y}{g}}=\sqrt{\frac{2*20}{9.8}}=2.02s$

$\text{Answer: }2.02s$