Question

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?

Answer 1

`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`