In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This give 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.
Due to air resistance, energy and velocity of particle keep on decreasing making the fall steeper shown in the figure.
When we neglect air resistance path is symmetric parabola $(OAB)$. When air resistance is considered path is asymmetric parabola $(OAC).$
For a given velocity, a projectile has the same range $R$ for two angles of projection if $t_1$ and $t_2$ are the times of flight in the two cases then
An object is projected in the air with initial velocity $u$ at an angle $\theta$. The projectile motion is such that the horizontal range $R$, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be $.......$ degree.
The position of a projectile launched from the origin at $t = 0$ is given by $\vec r = \left( {40\hat i + 50\hat j} \right)\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take $g = 10\, ms^{-2}$)
A projectile is thrown at an angle $\theta$ with the horizontal and its range is $R_1$. It is then thrown at an angle $\theta$ with vertical and the range is $R_2$, then
A stone is projected from the ground with velocity $25\,m/s$. Two seconds later, it just clears a wall $5 \,m$ high. The angle of projection of the stone is ........ $^o$ $(g = 10m/{\sec ^2})$