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??

Question

Charlotte Aaron · Accepted Answer

Maximum height after release is achieved in approximately

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

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")#

Nora Arzate · Answer

undefined The time it takes for a projectile to reach its maximum height can be calculated using the formula:

$$ t_{\text{max}} = \frac{v_y}{g} $$

where:
- $ v_y $ is the initial vertical component of velocity
- $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $)

Given that the projectile is shot at an angle of $ \frac{\pi}{6} $ radians (which is equivalent to $ 30^\circ $), and the initial velocity is $ 28 \, \text{m/s} $, the vertical component of the initial velocity ($ v_y $) can be calculated using trigonometric functions:

$$ v_y = v \cdot \sin(\theta) $$

$$ v_y = 28 \cdot \sin\left(\frac{\pi}{6}\right) $$

$$ v_y = 28 \cdot \frac{\sqrt{3}}{2} $$

$$ v_y = 14\sqrt{3} $$

Now, we can calculate the time it takes for the projectile to reach its maximum height:

$$ t_{\text{max}} = \frac{v_y}{g} $$

$$ t_{\text{max}} = \frac{14\sqrt{3}}{9.8} $$

$$ t_{\text{max}} ≈ 1.517 \, \text{seconds} $$

Therefore, the projectile will reach its maximum height approximately $ 1.517 $ seconds after being shot.