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.

Question

Eshan · Answer

Dear student,

The horizontal and vertical componetns of veloicty are

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

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

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

Vertical component of velocity= $v_y=u_y-gt$

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

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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.

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.

1 Answers

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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.

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.

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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.