Question

A body is projected with an initial velocity 20 m/s at 60 to the horizontal the displacement after 2 seconds is

Answer 1

Answer:

Displacement is 24. 786

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Answer 2

Answer:

The displacement after 2 seconds is $(20)\hat{i}+(20\sqrt{3}+20)\hat{j}$.

Explanation:

The velocity with which the body is projected=20m/s

Then,

The velocity along the x-axis is,

$v_x=20cos60\textdegree=10m/s$       (1)

The velocity along the y-axis is,

$v_y=20sin60\textdegree=20\times \frac{\sqrt{3}}{2}=10\sqrt{3} m/s$      (2)

Now, for displacement after 2 seconds for both x and y direction is given as,

Along x-axis

$S_x=v_xt$          (3)

By substituting the value of $v_x$ and $t$ in equation (3) we get;

$S_x=10\times 2=20m$

Along y-axis

$S_y=v_yt+\frac{1}{2}gt^2$         (4)

g=acceleration due to gravity=10m/s²

By substituting the value of $v_y$,$t$, and g in equation (4) we get;

$S_y=10\sqrt{3}\times 2+\frac{1}{2}\times 10\times 2^2=20\sqrt{3}+20m$

The net displacement of the particle is,

$S=S_x+S_y$             (5)

By substituting the equations $S_x$ and $S_y$  in equation (5) we get;

$S=(20)\hat{i}+(20\sqrt{3}+20)\hat{j}$

Hence, the displacement after 2 seconds is $(20)\hat{i}+(20\sqrt{3}+20)\hat{j}$.