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amistre64

  • 3 years ago

Worst worded question to date: A brick is thrown upward from the top of a building at an angle of 25degrees to the horizontal and with an initial velocity of 15m/s. If the brick is in flight for 3 seconds, how tall is the building?

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  1. amistre64
    • 3 years ago
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    It took me a little bit to assume that the 3 seconds in flight was a terminanting time, and not just a random interval :)

  2. amistre64
    • 3 years ago
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    |dw:1346854922454:dw|

  3. amistre64
    • 3 years ago
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    \[-4.9(3)^2+15(3)sin(25^o)+h_o=0\]

  4. ash2326
    • 3 years ago
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    Could I suggest something? @amistre64 sir

  5. amistre64
    • 3 years ago
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    of course

  6. ash2326
    • 3 years ago
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    only the velocity in vertical direction is of use here \[V_v= 15 \sin 25 \] now let's find the time it takes the velocity in vertical direction to get to 0 and the distance it covers|dw:1346855419638:dw| let the height be h1 and time be t1 here acceleration due to gravity=10 m/s^2 in downwards direction so -10 \[v-u=at\] \[0-V_v=-10*t_1\] we could find t_1 from here \[v^2-u^2=2as\] \[0-(V_V)^2=2*-10*h_1\] h1 from here now the remaining time it takes to reach ground 3-t1 using this we could find h2 building's height is h2-h1

  7. ash2326
    • 3 years ago
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    for h2 \[s=ut+1/2 at^2\] \[h_2=0\times (3-t_1)+\frac 12 (-10) (3-t_1)^2\]

  8. amistre64
    • 3 years ago
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    while i can understand the concepts you presented, it does seem like an awful lot of extra workings. I am a mathematician at heart, and by default very lazy :)

  9. ash2326
    • 3 years ago
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    :) he he

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