kanwal32
  • kanwal32
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
Physics
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kanwal32
  • kanwal32
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
Physics
chestercat
  • chestercat
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kanwal32
  • kanwal32
kanwal32
  • kanwal32
@Michele_Laino pls help in hurry i want to solve this doubt
kanwal32
  • kanwal32
is there any way

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Michele_Laino
  • Michele_Laino
in general I solve that type of problem using vectors equations, namely: \[\large \left\{ \begin{gathered} {\mathbf{OP}}\left( t \right) = {\mathbf{O}}{{\mathbf{P}}_{\mathbf{0}}} + {{\mathbf{v}}_{\mathbf{0}}}t + \frac{1}{2}{\mathbf{g}}{t^2} \hfill \\ {\mathbf{v}}\left( t \right) = {{\mathbf{v}}_{\mathbf{0}}} + {\mathbf{g}}t \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
Then I consider a reference frame and I will rewrite those equations in that reference frame
Michele_Laino
  • Michele_Laino
for example, if we pick the subsequent reference frame: |dw:1433609868699:dw|
Michele_Laino
  • Michele_Laino
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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