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
plz send me the answer
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.
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
To find the value of 9 sin²(α), we need to first understand the concept behind the maximum angle at which a stone can be thrown such that its distance from the thrower is always increasing.
Let's consider a situation where a stone is thrown with an initial velocity at an angle α to the horizontal. The trajectory of the stone will be a parabolic curve. As the angle increases, the range (horizontal distance covered by the stone) will also increase until it reaches a maximum value.
Now, for the stone's distance from the thrower to always increase, we need to ensure that the stone never falls back to the ground. This means that the stone's vertical component of velocity should never become zero or negative.
The vertical component of velocity can be represented as v * sin(α), where v is the magnitude of the initial velocity. For the distance to always increase, this vertical component of velocity should always be positive.
Therefore, we can write the condition as:
v * sin(α) > 0
Since v is a positive quantity (magnitude of velocity), we can simplify the condition to:
sin(α) > 0
Now, the maximum angle at which sin(α) is positive is 90 degrees. Any angle between 0 and 90 degrees will satisfy sin(α) > 0. Therefore, the maximum angle (α) is 90 degrees.
Finally, substituting α = 90 degrees into the expression 9 sin²(α), we get:
9 sin²(90°) = 9 * 1² = 9
Hence, the value of 9 sin²(α) is 9.