1) Horizontal is same for angle of projection θ and (90-θ).
2) a shell is fired at an angle 30° to the horizontal with a velocity 196 m/s. Find the total time of flight, horizontal range and maximum height attained?

2
can u xplin it
nope
i ma sorry i have to go
we know that
range=(u)square sin 2tita /g
then according to question,
range=(196)(196)( sin 2(30))/10 we know g=10
=38416 (root3)/2 /10
=38416(1.7320)/20
t=2u sin(tita)/g
=2(196)sin(30)/10
=392(1/2) /10
h max=(u) square( sin)
square (tita)/2g
=(196)square (sin)square(30)/2(10)
=38416(1/2)(1/2)/20

Answers

2014-11-04T14:08:42+05:30
We know that
range=(u)square sin 2tita /g
then according to question,
range=(196)(196)( sin 2(30))/10  we know g=10
        =38416 (root3)/2 /10
       =38416(1.7320)/20     
 t=2u sin(tita)/g
 =2(196)sin(30)/10
 =392(1/2) /10
h max=(u) square( sin)
square (tita)/2g 
   =(196)square (sin)square(30)/2(10)
  =38416(1/2)(1/2)/20
0
2014-11-04T14:19:45+05:30

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Angle of projection (α) = 30
velocity (u) = 196 m/s

total time of flight =  \frac{2usin( \alpha )}{g} = \frac{2*196*sin(30)}{9.8} =\frac{2*196*0.5}{9.8} =20s

horizontal range =  \frac{ u^{2}*sin(2 \alpha ) }{g} = \frac{ 196^{2}*sin(60) }{9.8} = \frac{ 196^{2}* \frac{ \sqrt{3} }{2} }{9.8} =3394.82m

maximum height =  \frac{ (u*sin(30))^{2} }{2*g} = \frac{ (196*0.5)^{2} }{2*9.8} = \frac{ 98^{2} }{19.6} =490m 
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