A projectile is shot from the ground at an angle of #pi/6 # and a speed of #1 m/s#. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?
1 Answer
H=0.012m and
Explanation:
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 dosplacement 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
By the problem
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