# A balll is projected with initial speed of 40 m/s making an angle of 30 degrees with the horizontal. Find(1) Vertical displacement(2) Horizontal displacement(3) velocity and angle it is making with horizontal at t=1 and t=3 sec.

Eshan
askIITians Faculty 2095 Points
4 years ago
Dear student,

The horizontal and vertical componetns of veloicty are

$\dpi{80} u_x=ucos30^{\circ}=\dfrac{\sqrt{3}}{2}u$
and$\dpi{80} u_y=usin30^{\circ}=\dfrac{1}{2}u$

Therefore vertical displacement in time 't’=$\dpi{80} u_yt-\dfrac{1}{2}gt^2$
Horizontal displacement in time ‘t’=$\dpi{80} u_xt$

Horizontal component of velocity after time ‘t’=$\dpi{80} v_x=u_x$(since no force acts in the horizontal direction)

Vertical component of velocity=$\dpi{80} v_y=u_y-gt$

Hence angle velocity forms with horizontal after time ‘t’=$\dpi{80} tan^{-1}(\dfrac{v_y}{v_x})=tan^{-1}(\dfrac{u_y-gt}{u_x})$