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At the top of its flight, it explodes, breaking into two parts, one of which has twice the mass of the other. The two particles land simultaneously. The lighter fragment lands back at launch point. Where does the other fragment land? What is the energy of explosion?

Velocity at top of flight $$=35\cos 60$$ $$=\frac{35}{2}m/s$$

Using momentum conservation $$(6)(\frac{35}{2})=(2)(\frac{-35}{2})+4v$$ $$v=35$$

Also distance from the origin at point of explosion $$d=\frac{u^2\sin2\theta}{2g}$$ $$d=\frac{245\sqrt 3}{4}$$ Distance covered by 4kg mass from the point of explosion in time $$T=\frac{2u\sin\theta}{2g}$$ $$=\frac{7\sqrt 3}{4}$$ distance covered $$x=\frac{245\sqrt 3}{4}$$

So total distance is $$d+x$$ $$\frac{245\sqrt 3}{2}$$ But the answer given is is 160m. What am I doing wrong?

Aditya
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    Your $d$ is twice the correct value. It takes $t_1=u\sin\theta/g\approx3.03$ seconds to reach the highest point of the trajectory. Therefore $$d=t_1 u\cos\theta\approx53.04\ \text{meters.}$$ The descent also takes time, but the lighter fragment has twice the horizontal velocity, so it cover the distance $x=2d=106$ meters. Total a bit under $160$ meters. – Jyrki Lahtonen Nov 26 '19 at 14:30
  • I got 3.03 seconds – Aditya Nov 26 '19 at 14:32
  • $3.03$ seconds $\times$ $17.5$ meters/second $=53$ meters. Or $d=245\sqrt{3}/8$ (using $g=10$ meters per second per second). – Jyrki Lahtonen Nov 26 '19 at 14:34
  • Should read the descent also takes time $t_1$ two comments up. Anyway, because the explosion doubled the horizontal speed of the lighter fragment, you should have $x=2d$. – Jyrki Lahtonen Nov 26 '19 at 14:38
  • Äsh, two more typos. Above "lighter fragment" should be replaced with "the forward going fragment" (=the heavier). – Jyrki Lahtonen Nov 26 '19 at 14:40
  • Shoudn't this be moved to physics stackexchange? – Aniruddha Deb Nov 26 '19 at 14:46
  • But isn’t $R=\frac{u^2\sin 2\theta}{g}$? I just halved the value to get $245\sqrt 3/4$ I see I am have got the wrong value, but why is it so? – Aditya Nov 26 '19 at 14:46
  • @AniruddhaDeb there is physics tag on Math SE specifically for such questions. – Aditya Nov 26 '19 at 14:47
  • I agree with @JyrkiLahtonen: There's a calculation error in the step where you calculated $d$: it should be $\frac{245 \sqrt{3}}{8}$, which on adding gives you the answer as $159.1275$, which is approximately 160 – Aniruddha Deb Nov 26 '19 at 14:55

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There's an error in the step where you obtain $d$ $$d = \frac{u^2 \sin{2\theta}}{2g} = \frac {1225 \times \sqrt{3}}{2\times 2 \times 10} = \frac{245 \sqrt{3}}{8}$$

on calculating $d+x$, we obtain $$d+x = \frac {245 \times 3 \sqrt 3}{8} \approx 160$$ which is the answer.

Aniruddha Deb
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