# 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