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kanwal32

  • one year ago

how to write the equation of projectile after it has reached its maximum height(suppose at time t1 and t2 there are at same height) without finding the the time taken to reach at it full height

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  1. kanwal32
    • one year ago
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    @Michele_Laino

  2. kanwal32
    • one year ago
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    @Michele_Laino pls help in hurry i want to solve this doubt

  3. kanwal32
    • one year ago
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    is there any way

  4. Michele_Laino
    • one year ago
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    in general I solve that type of problem using vectors equations, namely: \[\large \left\{ \begin{gathered} {\mathbf{OP}}\left( t \right) = {\mathbf{O}}{{\mathbf{P}}_{\mathbf{0}}} + {{\mathbf{v}}_{\mathbf{0}}}t + \frac{1}{2}{\mathbf{g}}{t^2} \hfill \\ {\mathbf{v}}\left( t \right) = {{\mathbf{v}}_{\mathbf{0}}} + {\mathbf{g}}t \hfill \\ \end{gathered} \right.\]

  5. Michele_Laino
    • one year ago
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    Then I consider a reference frame and I will rewrite those equations in that reference frame

  6. Michele_Laino
    • one year ago
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    for example, if we pick the subsequent reference frame: |dw:1433609868699:dw|

  7. Michele_Laino
    • one year ago
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    then those vector equation can be rewritten as follows: \[\Large \begin{gathered} \left\{ \begin{gathered} z\left( t \right) = {z_0} + {v_0}\left( {\sin \theta } \right)t - \frac{1}{2}g{t^2} \hfill \\ x\left( t \right) = {x_0} + {v_0}\left( {\cos \theta } \right)t \hfill \\ \end{gathered} \right. \hfill \\ \hfill \\ \left\{ \begin{gathered} {v_z}\left( t \right) = {v_0}\left( {\sin \theta } \right) - gt \hfill \\ {v_x}\left( t \right) = {v_0}\left( {\cos \theta } \right) \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \] |dw:1433610127064:dw|

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