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??

1 Answer
May 4, 2016

Answer:

Maximum height after release is achieved in approximately

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

Explanation:

Tony B

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")#