Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `12 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?

Answer 1

The distance is `=6.36m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=12*sin(1/6pi)`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=12sin(1/6pi)-g*t`

`t=12/g*sin(1/6pi)`

`=0.61s`

Resolving in the horizontal direction `rarr^+`

To find the horizontal distance, we apply the equation of motion

`s=u_x*t`

`=12cos(1/6pi)*0.61`

`=6.36m`