Question

A projectile is shot at a velocity of `8 m/s` and an angle of `pi/6`. What is the projectile's peak height?

Answer 1

Equations of motion are used to find an answer of 0.816 m

Explanation:

Use the equation of motion that reads:

`v^2` = `u^2` - `2gy`

Where `v` is the final vertical velocity, `u` is the initial vertical velocity and `y` is the change in altitude (the vertical displacement) of the projectile.

Now, the launch angle is `pi`/6, so `u` = 8 sin `pi/6` = 8 (0.5) = 4.0 `m/s`

"Reaches its greatest height" means `v` = 0, so putting it together, we get

0 = `4.0^2` - 2 (9.8)`y`

`y` = `16/ 19.6` = 0.816 m