Question

A particle is projected from ground with velocity $40\sqrt{2}m/sat{45}^{\circ }.$ Find the displacement of the particle after 2 s. $(g=10m/s^2)$

Answer 1

Given : $u=40\sqrt{2}m/s$ $t=2$ s $a_y=-g=-10m/s^2$

$\therefore$ $u_x=ucos{45}^o=40m/s$ and $u_y=usin{45}^o=40m/s$

x direction : $S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$ where $t=2$ s and $a_x=0$

$\therefore$ $S_x=40\times 2+0=80$ m

y direction : $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$ where $t=2$ s

$\therefore$ $S_y=40\times 2-\frac{10\times 2^2}{2}=60$ m

$⟹$ Displacement of the particle $S=\sqrt{S_x^2+S_y^2}=\sqrt{{80}^2+{60}^2}=100$ m