Question

If a projectile is shot at an angle of `pi/6` and at a velocity of `28 m/s`, when will it reach its maximum height??

Answer 1

Maximum height after release is achieved in approximately

`~~1.427" seconds to 3 decimal places "color(red)( larr "corrected value")`

Explanation:

Note that `sin(pi/6)=1/2`

Let upwards velocity be positive and due to initial projectile force
Let downward velocity be negative and due to gravity
Let time in seconds be `t` and time at at any moment be `t_i`
Let time at maximum height be `t_m`
Let the unit second be represented as `s`
Let the unit distance be represented by `m`

Acceleration due to gravity is `9.81 m/s^2`

Assumption: there is no drag or any other forces involved

The maximum height is when upward velocity equals downward velocity

Downward velocity at any instant `i -> 9.81t_i`

`color(red)("Correcting an omission "->xx [sin(pi/6)=1/2])`

Maximum height is achieved at `" "color(red)(1/2xx)28 m/s-9.81 m/s^2xxt_m s=0`

Thus `" "28/(color(red)(2))color(white)(.) m/s=9.81color(white)(.) m/s^2xxt_m s`

`color(green)("Did you know you can treat units in the same way that you treat algebra?")`

`=>t_m s=(color(red)(14))/9.81 ->_("units")cancel(m)/(cancel(s))xxs^(cancel(2))/(cancel(m)`

`color(blue)(=> t_m~~1.427" seconds to 3 decimal places")`