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

Answer 1

plz send me the answer

Answer 2

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.

Answer 3

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