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.



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