Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : DEC 2015

Horizontal ( +ve) Vertical ( +ve)

Initial velocity $\hspace{1cm}V_H=12 \sin 60 m/s \hspace{1cm} u_V=12 \cos60 m/s$

Acceleration $\hspace{1cm}a_H= 0 \hspace{1cm} a_V=g=9.81m/s^2$

Displacement $\hspace{1cm} x=11.3m \hspace{1cm} y=H$

Final velocity $\hspace{1cm} V_H=12 \sin 60 m/s\hspace{1cm} v_V=?$

Time $\hspace{1cm}t=? \hspace{1cm}t=?$

Horizontal motion:-

Using $s=ut+ \dfrac 12 at^2,$ we get,

$\hspace{1cm} x=V_H t [\because a_H=0] \\ t = \dfrac x{V_H} = \dfrac {11.3}{12\sin60} \\ t=1.087s$

Vertical motion:-

Using $s=ut + \dfrac 12 at^2,$ we get,

$H=u_V t + \dfrac 12 gt^2=(12\cos 60)(1.087)+ \dfrac 12 (9.81)(1.087)^2 \\ H=12.323m \Rightarrow Ans$