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khanitha

  • 4 years ago

A piece of marble is projected from earth's surface with velocity of 50 m/s. 2 seconds later, it just clears a wall 5 m high. What is the angle of projection

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  1. Awstinf
    • 4 years ago
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    Maybe you could make a triangle out of it so 2 seconds it barely clears the wall so the hyp is 100m and the opposite side from the angle is 5m and use inverse sin to find the angle so oppo/hyp 5/100, I got 2.87 degrees that seems a little small though...

  2. Awstinf
    • 4 years ago
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    |dw:1333208034298:dw| heres a picture

  3. Awstinf
    • 4 years ago
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    apparently its 60 degrees says another site

  4. salini
    • 4 years ago
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    @awstinf u totally ignord gravity acting downward(it gives the body accelerated motion)not uniform

  5. salini
    • 4 years ago
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    so wat u do is in horizontal direction,the cmponent of velocity is 50cos@ in time 2 seconds it will cover a distance 50cos@*2 |dw:1333208698171:dw| tan@=sin@/cos@=100/50cos@ solve it

  6. amistre64
    • 4 years ago
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    shouldnt there be a distance part to this?

  7. salini
    • 4 years ago
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    |dw:1333208865379:dw| sry i made a mistake in the diagram

  8. salini
    • 4 years ago
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    ?

  9. amistre64
    • 4 years ago
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    lol, i spose we are doing: 2(50) cos(a) to determine a distance from the wall :)

  10. amistre64
    • 4 years ago
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    \[h(t)=-\frac{g}{2}t^2+V_o sin(a)\ t+H_o\] \[h(2)=5=-9.8+2(50)sin(a)+0\] \[5+9.8=100sin(a)\] \[\frac{14.8}{100}=sin(a)=.1480\] \[a=sin^{-1}(.1480)\] maybe

  11. salini
    • 4 years ago
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    something has caught me! i am reading all the question wrong1 |dw:1333209509913:dw| use 5m instead of 100m that gives tan@=sin@/cos@=5/100cos@ sin@=1/20=0.05

  12. salini
    • 4 years ago
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    amistre ur approach is absolutely right check ur 9.8t^2 part u gobbled up the t^2

  13. amistre64
    • 4 years ago
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    :) yeah, i seen something was amiss, but couldnt place it. thanx

  14. hosein
    • 4 years ago
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    Hi use this formula\[y=(-1/2)g t^2+v _{0}\sin \theta+y _{0}\] if you take y=5,g=9.81,t=2,v0=50,y0=0 theta yields29.49 degrees

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