Manipal Entrance Test 2015 Physics Paper

At the two points of the trajectory during projectile motion, the horizontal component of velocity is same. Then,
150×

1

2

=v×

1

√2

⇒v=

150

√2

ms−1
Initially,
uy=u‌sin‌60=

150√3

2

ms−1
Finally, vy=v‌sin‌45° =

150

√2

×

1

√2

=

150

2

ms−1
But vy=uy+ayt
⇒

150

2

=

150√3

2

−10t
⇒10t=

150(√3−1)

2

⇒t=7.5(√3−1)s