June 13, 2008

Q. 1.10 - "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$:

Two bodies were thrown simultaneously from the same point; one, straight up, and the other, at an angle of $\theta = 60^o$to the horizontal. The initial velocity of each body is equal to $v_0 = 25m/s$. Neglecting the air drag, find the distance between the bodies at $t = 1.70s$later.

$\textbf{Answer}$:
We will use vector notation for this problem. Let $\vec{j}$denote the upward direction and $\vec{i}$denote the horizontal direction towards which the second object was thrown.

Since the first object was thrown vertically, its position vector after $t$is = $(v_0 t - 1/2 gt^2)\vec{j}$.

The second object was thrown with its initial velocity as $v_0\cos\theta\vec{i} + v_0\sin\theta\vec{j}$. Since there is no horizontal force acting on the second body, its horizontal position after time $t$is $v_0 t\cos\theta\vec{i}$. And its vertical position at this time is $(v_0 t\sin\theta - 1/2gt^2)\vec{j}$

The displacement vector between the two bodies is =

\begin{displaymath}v_0 t\cos\theta\vec{i} + v_0 t(\sin\theta - 1)\vec{j} \end{displaymath}

Hence the distance between the two bodies is =

\begin{displaymath}\sqrt {(v_0 t)^2 \cos^2\theta + (\sin\theta - 1)^2} \end{displaymath}

\begin{displaymath}= v_0 t \sqrt { 2 - 2\sin\theta } \end{displaymath}

\begin{displaymath}= v_0 t \sqrt { 2 (1 - \sin\theta) } \end{displaymath}

Using values, the distance = $ 25 \times 1.7 \sqrt {2 (1 - \frac{\sqrt{3}}{2})} $= $42.5 \sqrt {0.268}$= 22 m

Posted by Apurva Sharan at 10:07 AM
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