when we calculate the time of flight for projectile with no drag we use second equation of motion i.e.
$$S=V_i t + at^2/2$$
further, we put $V_i=V_i\sin\theta$ and $a=-g$ i.e.:
$$S=V_i t\sin\theta – gt^2/2$$
further if we solve this equation it becomes $T= 2V_i \sin\theta/g$
from this formula it seems that this equation only calculate the time for vertical component of projectile (because $V_i\cos\theta$ is ignored). So it means that if a projectile is thrown at an angle this equation will not provide us an accurate information about the time spent by the projectile from projection point to the impact point! (is total time of flight is something different than what i thought?)
[Physics] Why we take only the vertical component of a projectile’s velocity while calculating the “Time Of Flight”
kinematicsprojectile
Best Answer
Neglecting the influences of air, the only things that affect time in the air are the initial vertical component of velocity and the downward acceleration due to gravity. The initial horizontal component of velocity only affects the horizontal distance the shell goes before hitting the ground.
Hope this helps.