Asked by jithin
                if the maximum angle to the horizontal at which a stone can be thrown such that its distance from thrower is always increasing is alfa ,then find the value of 9 sin square alfa
            
            
        Answers
                    Answered by
            jithin
            
    plz send me the answer
    
                    Answered by
            oobleck
            
    if the stone is thrown with velocity v at an angle α, then
vertical distance = v sinα t - 4.9t^2
horizontal distance = v cosα t
So, the distance z from the thrower is
z = √((v sinα t - 4.9t^2)^2 + (v cosα t)^2)
z = t√(24.01t^2 - 9.8v sinα t + v^2)
dz/dt = √(48.02t^2 - 14.7v sinα t + v^2)/√(24.01t^2 - 9.8v sinα t + v^2)
you want dz/dt to always be positive.
Note that the denominator is always positive, since the discriminant is always negative. So you only have to worry about the numerator.
See what you can do with that.
    
vertical distance = v sinα t - 4.9t^2
horizontal distance = v cosα t
So, the distance z from the thrower is
z = √((v sinα t - 4.9t^2)^2 + (v cosα t)^2)
z = t√(24.01t^2 - 9.8v sinα t + v^2)
dz/dt = √(48.02t^2 - 14.7v sinα t + v^2)/√(24.01t^2 - 9.8v sinα t + v^2)
you want dz/dt to always be positive.
Note that the denominator is always positive, since the discriminant is always negative. So you only have to worry about the numerator.
See what you can do with that.
                    Answered by
            Anonymous
            
    I wonder if the question is phrased correctly.
Perhaps they mean horizontal range.
x = u t = (Vi cos a) t
y = 0 + (Vi sin a) t - 4.9 t^2
hits ground when y = 0
4.9 t^2 = (Vi sin a)t
it is on the ground when t = 0 of course so look at the other solution
t = (Vi/4.9) sin a
then
x = (Vi cos a )(Vi/4.9) sin a = (Vi^2/4.9) sin a cos a = (Vi^2/4.9)(1/2) sin 2a
sin 2a is max when a = 45 degrees
then
9 sin^2 45 deg = 9 (1/2) =4.5
    
Perhaps they mean horizontal range.
x = u t = (Vi cos a) t
y = 0 + (Vi sin a) t - 4.9 t^2
hits ground when y = 0
4.9 t^2 = (Vi sin a)t
it is on the ground when t = 0 of course so look at the other solution
t = (Vi/4.9) sin a
then
x = (Vi cos a )(Vi/4.9) sin a = (Vi^2/4.9) sin a cos a = (Vi^2/4.9)(1/2) sin 2a
sin 2a is max when a = 45 degrees
then
9 sin^2 45 deg = 9 (1/2) =4.5
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