## 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

1. kanwal32

@Michele_Laino

2. kanwal32

@Michele_Laino pls help in hurry i want to solve this doubt

3. kanwal32

is there any way

4. Michele_Laino

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

5. Michele_Laino

Then I consider a reference frame and I will rewrite those equations in that reference frame

6. Michele_Laino

for example, if we pick the subsequent reference frame: |dw:1433609868699:dw|

7. 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|