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

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