# 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??

May 4, 2016

Maximum height after release is achieved in approximately

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

#### Explanation: Note that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

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 $9.81 \frac{m}{s} ^ 2$

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 $i \to 9.81 {t}_{i}$

$\textcolor{red}{\text{Correcting an omission } \to \times \left[\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}\right]}$

Maximum height is achieved at $\text{ } \textcolor{red}{\frac{1}{2} \times} 28 \frac{m}{s} - 9.81 \frac{m}{s} ^ 2 \times {t}_{m} s = 0$

Thus $\text{ } \frac{28}{\textcolor{red}{2}} \textcolor{w h i t e}{.} \frac{m}{s} = 9.81 \textcolor{w h i t e}{.} \frac{m}{s} ^ 2 \times {t}_{m} s$

$\textcolor{g r e e n}{\text{Did you know you can treat units in the same way that you treat algebra?}}$

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

$\textcolor{b l u e}{\implies {t}_{m} \approx 1.427 \text{ seconds to 3 decimal places}}$