Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `3 /2 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?

Answer 1

`s approx 0.099 m`

Explanation:

Step 1: Maximum height will be reached when `sin theta` component of projectile's velocity becomes 0 (zero)

Use the formula
`v=u+g t`
Given `v=0, u= 3/2 . sin (pi / 6)`; as `theta=pi/6`
Take `g= 9.81 m/s^2`
Since gravity is acting against the velocity so it is deceleration and `-` sign is used in front of g
`t= (3/2 . sin (pi / 6))/9.81 s`

Step 2:Distance traveled horizontally in time t is due to `cos theta` component of the initial velocity of the projectile. Assuming zero air friction, and using the equation

`s=ut`
`s= 3/2 . cos (pi / 6) times (3/2 . sin (pi / 6))/9.81 m`