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Aadarsh

  • 2 years ago

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)

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  1. Aadarsh
    • 2 years ago
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    @Mani_Jha , @Ishaan94 please help.

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

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

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

  5. srinidhijha
    • 2 years ago
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    k

  6. apoorvk
    • 2 years ago
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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'.

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

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

  9. Aadarsh
    • 2 years ago
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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?

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

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

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

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

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

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

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

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

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

  19. srinidhijha
    • 2 years ago
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    look the pa

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  20. Aadarsh
    • 2 years ago
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    Thanks apoorv bhai.

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

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

  23. Aadarsh
    • 2 years ago
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    samjha dijiye

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

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  25. srinidhijha
    • 2 years ago
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    samajh aaya?

  26. srinidhijha
    • 2 years ago
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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

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

  28. srinidhijha
    • 2 years ago
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    k good

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

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

  31. srinidhijha
    • 2 years ago
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    ha ha

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