Question

A body is projected from height of 60 m with a velocity 10 m/s at angle 30 degrees to horizontal . Time of flight of body in (g=10 m/s2

Answer 1

Answer:

h=60 m (downward) (+ve), uy

=10sin30o

(initial vertical velocity)

time of light will be time to reach ground i.e. to cover

60 m displacement / ↑−ve & ↓+ve

Put h=60 m,u→4u=− 10/2

=−5 m/s

9=10 m/s2

We get

60=−5t+5t2 or t2 −t−12=0 or t2

−2t+3t−12=0 or t(t−4)+3(t−4)=0

⇒ t=4 sec

hope it will help you....