If a projectile is shot at an angle of #pi/6# and at a velocity of #24 m/s#, when will it reach its maximum height??

1 Answer
Jun 16, 2016

Answer:

#t~~1.223 " seconds"# to 2 decimal places

Note that #9.81# for g is an approximate figure so the solution is an approximation to close limits.

Explanation:

Tony B
Let time in seconds be #t#
Let vertical height by #h#

Vertical height at any time is

#h=[24sin(pi/6)xxt] - [1/2xx9.81xxt^2]#

The #ul("rate of change of h")# will be 0 at maximum height

#=> (dh)/(dt)~~ 24sin(pi/6)-9.81t#

But at maximum height #(dy)/(dt)=0# giving

#9.81t~~24sin(pi/6)#

#=> t=24/9.81sin(pi/6)#

But #sim(pi/6)=1/2#

#=>t~~12/9.81#

#t~~1.223 " seconds"# to 2 decimal places