Here we have a question of a 2 dimensional movement. I know that it is needed to get its second derivatives for acceleration but then what should I do?
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$\begingroup$ Do you know how to use a scalar product? $\endgroup$– CrossCommented Jan 7, 2023 at 16:15
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1$\begingroup$ Yes but I get really confused to how use it in this situation. $\endgroup$– OriCommented Jan 7, 2023 at 16:16
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$\begingroup$ Find $\vec a$ and $\vec r$, and then you can get your answer. $\endgroup$– CrossCommented Jan 7, 2023 at 16:19
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1$\begingroup$ I love the instruction “Take g=10 m/s$^2$”. $\endgroup$– GhosterCommented Jan 7, 2023 at 21:38
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Your position vector is: $$ \vec{r}(t)=\big( x(t),y(t) \big) $$ and your acceleration vector is: $$ \vec{a}(t) = \frac{d^2\vec{r}}{dt^2} = \big( \ddot{x}(t), \ddot{y}(t) \big) $$ then, the internal product is: $$ \vec{r} \, \cdot \vec{a} = |\vec{r}| \, |\vec{a}| \,cos(\theta) $$ but also: $$ \vec{r} \, \cdot \vec{a} = x\ddot{x} + y\ddot{y} $$ therefore: $$ \theta= cos^{-1} \bigg( \dfrac{x\ddot{x} + y\ddot{y}}{|\vec{r}| \, |\vec{a}|} \bigg) $$
