Question

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

Answer 1

initialverticalvelocity$=usin45°=40\sqrt{2}\left[\frac{1}{\sqrt{2}}\right]=40m/s$

constanthorizontalvelocity$=ucos45°=40\sqrt{2}\left[\frac{1}{\sqrt{2}}\right]=40m/s$

after2seconds:verticaldisplacement$=s=ut+\frac{1}{2}at²=80-4.905\left(4\right)=80-19.62=60.38m$

horizontaldisplacement=speedxtime$=40\times 2=80m$

$sᵣₑ=\sqrt{\left[\right(80)²+(60.38\left)²\right]}=100.23m$

at anangle of $tanˉ¹\frac{60.38}{80}=37.044°$above horizontal