Question

A particle is projected from ground with velocity $40\sqrt{2}$ m/s at ${45}^o$. Find velocity.

Answer 1

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

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

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

horizontal displacement =speedxtime$=40\ast 2=80m$

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

atanangleof$tanˉ¹\frac{60.38}{80}=37.044°$abovehorizontal