We have a ball projected at 20 m/s at an angle $\pi/6$ from the $x$-axis. Now we have to calculate $x$-coordinate of the ball when it's velocity is perpendicular to projection velocity.
I am thinking about how to solve this problem. Is it related to the equation of trajectory? What I have figured out so far, is that it will be making angle $\pi/3$ from $x$-axis at this instant how to get the $x$-coordinate?
Best Answer
You can start off by making components of initial velocity, the component perpedicular to ground will provide flight and that along the ground will take the projectioe forward.
I assume y direction is vertical and x is ground
So the velocity perendicular to ground (velocity × sine of angle from ground) is 10 m w/s. The acceleration of the body is that of gravity and is takenweas 10 m/s2 for simplicity. Now you can calculate the time for when the velocity would become perpendicular to original velocity ( it would be when the body will be coming down with velocity at 60 degrees from original velocity) we can find the y component of velocity at any time with : v = u + at .
Using trigonometry and above equatiob you can find that the projectile would take 4 seconds to get to such a position. But by calculating time of flight of projectile you will find that it is only 2 seconds. Hence the projectile would not come to a point where its velocity may be perpendicular to its launch velocity.