A projectile is shot from the ground at an angle of #( pi)/3 # and a speed of #6 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?
1 Answer
Let the velocity of projection of the object be u with angle of projection
The vertical component of the velocity of projection is
Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical displacement h will be zero. So applying the equation of motion under gravity we can write
where
The horizontal displacement during this T sec is
The time t to reach at the peak is half of time of flight (T)
So
The horizontal displacement during time t is
If H is the maximum height then
So the distance of the object from the point of projection when it is on the peak is given by