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In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is include? Sketch such a trajectory and explain why you have drawn it that way.
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When air resistance acts on projectile then its vertical and horizontal then particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper than rise as shown in the figure.
When we are neglecting air resistance path was symmetric parabola $(O A B)$. When air resistance is considered path is asymmetric parabola $(\bar{O} A \bar{C})$.
By formula $R=\frac{u^2}{g} \sin 2 \theta$, and $H_{\max }=\frac{u^2 \sin ^2 2 \theta}{2 g}$
So, $\left(h_1 < h_2\right)$ and $\left(R_1 < R_2\right)$.
But time of flight for both will remain same.
When we are neglecting air resistance path was symmetric parabola $(O A B)$. When air resistance is considered path is asymmetric parabola $(\bar{O} A \bar{C})$.
By formula $R=\frac{u^2}{g} \sin 2 \theta$, and $H_{\max }=\frac{u^2 \sin ^2 2 \theta}{2 g}$
So, $\left(h_1 < h_2\right)$ and $\left(R_1 < R_2\right)$.
But time of flight for both will remain same.
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