Question

A particle is projected from ground with velocity 40 m/s at ${60}^{\circ }$ with horizontal. Find speed of particle when its velocity is making ${45}^{\circ }$ with horizontal. Also find the times (s) when it happens. $(g=10m/s^2)$

Answer 1

Given : $u=40m/s$ $a_y=-g=-10m/s^2$

$\therefore$ $u_x=ucos{60}^o=20m/s$ and $u_y=usin{60}^o=34.64m/s$

x direction : $a_x=0$ $⟹V_x=u_x=20m/s$

Also $\frac{V_y}{V_x}=tan{45}^o=1$ $⟹V_y=V_x=20m/s$

y direction : $V_y=u_y+a_{y}t$

Upward motion : $20=34.64-10t_1$ $⟹t_1=1.464$ s

Downward motion : $-20=34.64-10t_2$ $⟹t_2=5.464$ s