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Aadarsh Group Title

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)

  • 2 years ago
  • 2 years ago

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  1. Aadarsh Group Title
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    @Mani_Jha , @Ishaan94 please help.

    • 2 years ago
  2. Aadarsh Group Title
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    @apoorvk , @srinidhijha, @AravindG please help

    • 2 years ago
  3. srinidhijha Group Title
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    what is angle of projection? is the question written as it is?

    • 2 years ago
  4. Aadarsh Group Title
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    Yeah, the question is as it is as I have written here.

    • 2 years ago
  5. srinidhijha Group Title
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    k

    • 2 years ago
  6. apoorvk Group Title
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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'.

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

    • 2 years ago
  8. apoorvk Group Title
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    LOL sorry - it would 'final' velocity, not time of flight - my bad lol.

    • 2 years ago
  9. Aadarsh Group Title
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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?

    • 2 years ago
  10. apoorvk Group Title
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    yeah, so use, (v-u)/g = t

    • 2 years ago
  11. srinidhijha Group Title
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    hey is the speed 97 ? u know the answers?

    • 2 years ago
  12. Aadarsh Group Title
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    Answer is 30 m/s

    • 2 years ago
  13. apoorvk Group Title
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    for a particle under free fall with initial velocity 'u=0'|dw:1338565657495:dw|

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

    • 2 years ago
  15. apoorvk Group Title
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    yeah, v=30, and height = 45 metres, right?

    • 2 years ago
  16. apoorvk Group Title
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    well, that is for a ground-to-ground projectile @Aadarsh

    • 2 years ago
  17. Aadarsh Group Title
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    @apoorvk bhai, ur answer is right.

    • 2 years ago
  18. Aadarsh Group Title
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    @apoorvk bhai, please show all the steps.

    • 2 years ago
  19. srinidhijha Group Title
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    look the pa

    • 2 years ago
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  20. Aadarsh Group Title
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    Thanks apoorv bhai.

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

    • 2 years ago
  22. Aadarsh Group Title
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    Yeh to samajh mein nahion aaya.....

    • 2 years ago
  23. Aadarsh Group Title
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    samjha dijiye

    • 2 years ago
  24. srinidhijha Group Title
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    see the attachment koi horizontal equal lene ki jarurat nh ihai

    • 2 years ago
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  25. srinidhijha Group Title
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    samajh aaya?

    • 2 years ago
  26. srinidhijha Group Title
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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

    • 2 years ago
  27. Aadarsh Group Title
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    Thakns srini bhaiiiiiiiii. samajh mein aa gaya.

    • 2 years ago
  28. srinidhijha Group Title
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    k good

    • 2 years ago
  29. apoorvk Group Title
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    Sri, great calligraphy! :P

    • 2 years ago
  30. AravindG Group Title
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    srry i am late :P

    • 2 years ago
  31. srinidhijha Group Title
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    ha ha

    • 2 years ago
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