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Aadarsh

A projectile is thrown horizontallly from the top of a tower and strikes teh ground after 3 second at an angle of 45 degrees with the horizontal. Find the height of the tower and speed with which the body was projected. (Take "g" = 9.8 m/s^2)

  • one year ago
  • one year ago

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  1. Aadarsh
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    @Mani_Jha , @Ishaan94 please help.

    • one year ago
  2. Aadarsh
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    @apoorvk , @srinidhijha, @AravindG please help

    • one year ago
  3. srinidhijha
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    what is angle of projection? is the question written as it is?

    • one year ago
  4. Aadarsh
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    Yeah, the question is as it is as I have written here.

    • one year ago
  5. srinidhijha
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    k

    • one year ago
  6. apoorvk
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    So when it strikes, basically the horizontal and vertical components of its velocity are equal (since it strikes the ground at 45 degrees) Now, let the initial velocity of projection be 'u' Now, the horizontal speed remains constant, but the vertical component increases from '0' to 'u', as it hits the ground. Now, vertical component (final) = v time of flight = sqrt(2gh) = 3 Find height, and then find 'u'.

    • one year ago
  7. Aadarsh
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    How time of flight = sqrt(2gh)? Can u show the complete steps and other equations also?

    • one year ago
  8. apoorvk
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    LOL sorry - it would 'final' velocity, not time of flight - my bad lol.

    • one year ago
  9. Aadarsh
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    Still i am not able to follow how is it equal to sqrt(2gh)? Can u show the workign of other steps also?

    • one year ago
  10. apoorvk
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    yeah, so use, (v-u)/g = t

    • one year ago
  11. srinidhijha
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    hey is the speed 97 ? u know the answers?

    • one year ago
  12. Aadarsh
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    Answer is 30 m/s

    • one year ago
  13. apoorvk
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    for a particle under free fall with initial velocity 'u=0'|dw:1338565657495:dw|

    • one year ago
  14. Aadarsh
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    Got it. Can we not say that:\[T = 2u \cos \theta / g = \sqrt{2}u/g\]

    • one year ago
  15. apoorvk
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    yeah, v=30, and height = 45 metres, right?

    • one year ago
  16. apoorvk
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    well, that is for a ground-to-ground projectile @Aadarsh

    • one year ago
  17. Aadarsh
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    @apoorvk bhai, ur answer is right.

    • one year ago
  18. Aadarsh
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    @apoorvk bhai, please show all the steps.

    • one year ago
  19. srinidhijha
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    look the pa

    • one year ago
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  20. Aadarsh
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    Thanks apoorv bhai.

    • one year ago
  21. apoorvk
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    See, ye samajh mein aaya ki horizontal and vertical components at the time of hitting the ground same honge?

    • one year ago
  22. Aadarsh
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    Yeh to samajh mein nahion aaya.....

    • one year ago
  23. Aadarsh
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    samjha dijiye

    • one year ago
  24. srinidhijha
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    see the attachment koi horizontal equal lene ki jarurat nh ihai

    • one year ago
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  25. srinidhijha
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    samajh aaya?

    • one year ago
  26. srinidhijha
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    isse pahle bhi ek attachment tha jisme U nikaala hai , dkho, Final velocity to 0 hi hoga naa jab wo gir jaaye to

    • one year ago
  27. Aadarsh
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    Thakns srini bhaiiiiiiiii. samajh mein aa gaya.

    • one year ago
  28. srinidhijha
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    k good

    • one year ago
  29. apoorvk
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    Sri, great calligraphy! :P

    • one year ago
  30. AravindG
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    srry i am late :P

    • one year ago
  31. srinidhijha
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    ha ha

    • one year ago
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