If a projectile is shot at an angle of #(7pi)/12# and at a velocity of #9 m/s#, when will it reach its maximum height?

1 Answer
Apr 17, 2017

Answer:

#0.63 sec#

Explanation:

We know that according to the law of motion .
#V=U+a*t#

#:.# this law is true for the frame of reference for non inertial frame of reference .

So it can be written as ,
#V_y=U_y+a_y*t#

we know that vertical component of the velocity is
#Usin theta#
here it can be written as

#Usin((7pi)/12)=Usin(pi-(3pi)/12)=Usin(pi/4)###as sin is positive in 2 quadrant

for attaining maximum height we get final velocity #V_y=0# so upon substituting the values in the equation we get .
#0=9*1/sqrt(2)-10t#

as the acceleration due to gravity is in negative direction.
so we get .

#9/(10*sqrt(2))=t#
#t=(9sqrt(2))/20=12.69/2=0.63 sec#