Question

a body is projected from ht.of 60m with a velocity 10m/s at angle 30 to the horizontal then time of flight of the body

Answer 1

Uniformly accelerated motion

Known:
h=-60 m ,because from above
vo=10 m/s
α=30⁰
Asked: time (t)?
Answer:
Use this formula:
h=voy.t-¹/₂gt²

we know that g=9,8 m/s² or 10 m/s² to make it more comfortable:

h=vo.sinα.t-¹/₂.gt²
-60=10.sin 30.t-¹/₂(10).t²
-60=10.(¹/₂).t-5t²
5t²-5t+60=0
  t²-t-12=0
(t-4) v (t+3)
t=4 s v t= -3 s

Or if you want to use g=9,8 m/s²,you have use ABC formula:
a=4,6
b=-5
c=60
$\boxed{x= \frac{-b\pm \sqrt{b^2-4ac} }{2a} } \\\ \bullet x_1=\frac{-b + \sqrt{b^2-4ac} }{2a} } \\\ \bullet x_2= \frac{-b - \sqrt{b^2-4ac} }{2a} }$

Time not negative ! So ,we use positive ,yes 4 sekon.



O-----------------------------OKAY----------------------------------O

Answer 2

Two ways...

1. find the time duration and the distance to reach the highest point.  Find the time duration to reach the ground...   Add the two durations.

         t1 = 10m/s * Sin30 / 10 m/s^2 = 1/2 sec
         h1 = (10 * sin 30)^2/ 2*10 = 1.25 m
         s = 1.25 + 60 m = 0 t + 1/2 * 10 * t2^2     =>  t2 = 3.5 sec
         t = t1 +t2 = 4 sec.

2.  Find the time duration to reach a height (negative vertical 
     displacement) of  - 60 m.
 
         s = u t + 1/2 g t^2
         - 60 = 10 * sin 30 t - 1/2 * 10 * t^2
         t^2 - t - 12 = 0
            t = 4 or -3 sec.
        so t = 4 sec.