When we analyse the projectile motion from any accelerated frame O as vecr_(o),vecu_(o) and veca_(o) respectively, express the following terms, vecr_(pO)+veca_(p)-vecr_(O),vecu_(pO)=vecu_(p)-vecu_(O) and veca_(pO)=veca_(p)-veca_(O) where P stands for projectile. Then using the following kinematical equations of the projectile (For constant acceleration) relative to the accelerating frame ,we have vecS_(pO)=vecu_(pO)t+1/2veca_(pO)t^(2),vecv_(pO) =vecu_(pO)+veca_(pO)t and v_(pO)^(2)=u_(pO)^(2)+2veca.vecs_pO Using the above expressions, anwer the following question:A projectile has initial velocity v_(0) realtive to the large plate which is moving with a constant upward acceleration a. Which of the following remains equal for the observers A and B?

When we analyse the projectile motion from any accelerated frame O as

1.	When we analyse the projectile motion from any accelerated frame O as vecr_(o),vecu_(o) and veca_(o) respectively, express the following terms, vecr_(pO)+veca_(p)-vecr_(O),vecu_(pO)=vecu_(p)-vecu_(O) and veca_(pO)=veca_(p)-veca_(O) where P stands for projectile. Then using the following kinematical equations of the projectile (For constant acceleration) relative to the accelerating frame ,we have vecS_(pO)=vecu_(pO)t+1/2veca_(pO)t^(2),vecv_(pO) =vecu_(pO)+veca_(pO)t and v_(pO)^(2)=u_(pO)^(2)+2veca.vecs_pO Using the above expressions, anwer the following question:A projectile has initial velocity v_(0) realtive to the large plate which is moving with a constant upward acceleration a. Which of the following remains equal for the observers A and B?
Answer» MAXIMUM height Range Time of flight ANGLE of projection Solution :Angle of projection remains EQUAL for both `A&B`.