A projectile is thrown from point P .it moves in such a way that it's distance from point P is always Increasing .find the maximum angle above horizontal with which is thrown .ignore air resistance
Answers
Suppose that the ball is projected with a velocity 'v' at an angle 'θ' to the horizontal.
The horizontal component of Velocity vector is v cosθ and the vertical component is v sinθ.
With this data, assuming the point P is at the origin, the position of the ball at any instant 't' is given by
Horizontal position = vt cosθ
Vertical position = vt sinθ - ½gt²
Distance of the ball (s) from P is
s = √ [ (vt cosθ)² + (vt sinθ - ½gt²) ]
s = √ [ v²t² cos² θ + v²t² sin²θ - vgt³ sinθ + 0.25g²t⁴ ]
s = √ [ v²t² - vgt³ sinθ + 0.25g²t⁴ ]
For the distance s to be an increasing function, always, the condition ds/dt > 0 should be satisfied.
ds/dt = ½ (2v²t - 3vgt² sinθ + g²t³) / √ [ v²t² - vgt³ sinθ + 0.25g²t⁴ ]
ds/dt = (v² - 1.5vgt sinθ + 0.5g²t²) / √ [ v² - vgt sinθ + 0.25g²t² ] > 0
0.5g²t² -1.5vgt sinθ + v² > 0
The LHS of the above inequality is a quadratic expression with variable 't'.
It is a property of any quadratic expression (at² + bt + c) that if the discriminant Δ = b² - 4ac < 0, then the sign of the value of the expression (at² + bt + c) and that of 'a' will always be the same.
In our expression, obviously, g² is always greater than zero. Thus, if, for the expression Δ<0 then, s will always be an increasing function,
(1.5vg sinθ)² - 4(0.5g²)v² < 0
sin²θ < 2/2.25
sinθ < 0.943
θ < Sin⁻¹(0.943)
θ < 70.53°
This is the maximum angle at which the ball can be projected.
I however 14 at first for same reasons others depict yet suppose the dad was his age at time of birth, wouldn't it be one more year before the youngster is 1? This implies when the child is 1, the dad is birth Age + 1.
I'd take current Age - 1 (37) and isolate by 2, which meets 18.5.
At that point subtract 5 years and equivalents 13.5.
I figure round down to 13 in absolute years. Furthermore, age is 13.