Question

what is time of flight of projectile whose launch angle is 30⁰ and initial velocity is 196 m/sec?​

Answer 1

Answer:

19.6

Explanation:

Time of flight for a body thrown at an angle from the ground [T] = 2u sin thita/g

where u is initial velocity of the body

and thita is the angle of projection

given, u=196m/s

  thita =30 degree

T={2*196*[1/2]}/10

 =196/10

 =19.6

Answer 2

20 sec

t is 2 u sin theta / g