Trajectory Optimization (6-23-2020)

Classical Mechanics Level 3

A particle is launched at time t = 0 t = 0 from ground level with initial speed v 0 v_0 at an angle θ \theta with respect to ground. Gravitational acceleration g g is downward.

Let T ( t ) T(t) be the particle's kinetic energy as a function of time, and let V ( t ) V(t) be the particle's gravitational potential energy as a function of time, measured with respect to ground. Let t f t_f be the time at which the particle lands.

What value of θ \theta (in degrees) makes the following true?

∫ 0 t f T ( t ) d t ∫ 0 t f V ( t ) d t = 1 \large{\frac{\int_0^{t_f} T(t) \, dt}{\int_0^{t_f} V(t) \, dt} = 1}

Note: 0 < θ < 9 0 ∘ 0 < \theta < 90^\circ


The answer is 60.0.

Steven Chase by Steven Chase , 37, USA

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2 solutions

For simplicity i am taking v 0 = v v_0=v
t f t_{f} is the total time of journey
0 = v sin ⁡ θ t f − 1 2 g t f 2 0=v\sin \theta t_{f}-\frac{1}{2} gt_{f}^{2}
solving the above equation leads to t f = 2 u sin ⁡ θ g t_{f}=\frac{2u \sin \theta}{g} write the expression for kinetic energy ( T ( t ) ) (T(t)) T ( t ) = 1 2 m ( v x 2 + v y 2 ) T(t)=\frac{1}{2} m(v_{x}^{2}+v_{y}^{2})
T ( t ) = 1 2 m ( v 2 cos ⁡ 2 θ + ( v sin ⁡ θ − g t ) 2 ) T(t)=\frac{1}{2} m(v^{2}\cos ^{2} \theta+(v\sin \theta -gt)^{2})
Lets evaluate ∫ 0 2 v s i n θ g T ( t ) d t = ∫ 0 2 v s i n θ g 1 2 m v 2 cos ⁡ 2 θ d t + 1 2 m ∫ 0 2 v s i n θ g ( v 2 sin ⁡ 2 θ + g 2 t 2 − 2 v sin ⁡ θ g t ) d t \int_{0}^{\frac{2vsin \theta}{g}} T(t) dt= \int_{0}^{\frac{2vsin \theta}{g}} \frac{1}{2} mv^{2}\cos ^{2} \theta dt+ \frac{1}{2}m \int_{0}^{\frac{2vsin \theta}{g}} (v^{2} \sin^{2} \theta+ g^{2}t^{2} -2v\sin \theta gt )dt
∫ 0 t f T ( t ) d t = m v 3 g ( sin ⁡ θ + 4 sin ⁡ 3 θ 3 − 2 sin ⁡ 3 θ ) \int_{0}^{t_{f}} T(t) dt=\frac{mv^{3}}{g} (\sin \theta+ \frac{4 \sin^{3} \theta }{3} -2\sin^{3} \theta)
Now,lets write the expression for potential energy ( V ( t ) ) (V(t)) V ( t ) = m g ( v sin ⁡ θ t − 1 2 g t 2 ) V(t)=mg(v\sin \theta t -\frac{1}{2}gt^{2})
Lets evaluate ∫ 0 2 v s i n θ g V ( t ) d t = 4 m v 3 sin ⁡ 3 θ 6 g \int_{0}^{\frac{2vsin \theta}{g}} V(t) dt = \frac{4mv^{3}\sin ^{3} \theta}{6g}
condition for a particular θ \theta ∫ 0 2 v s i n θ g T ( t ) d t = ∫ 0 2 v s i n θ g V ( t ) d t \int_{0}^{\frac{2vsin \theta}{g}} T(t) dt = \int_{0}^{\frac{2vsin \theta}{g}} V(t) dt


m v 3 g ( sin ⁡ θ + 4 sin ⁡ 3 θ 3 − 2 sin ⁡ 3 θ ) = 4 m v 3 sin ⁡ 3 θ 6 g \frac{mv^{3}}{g} (\sin \theta+ \frac{4 \sin^{3} \theta }{3} -2\sin^{3} \theta) = \frac{4mv^{3}\sin ^{3} \theta}{6g}
solving the above equation leads to in degrees θ = 60 \boxed{\theta = 60}

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I will be happy if anyone will post the solution of this problem by solving it with python programming.
Thanks in advance

A Former Brilliant Member - 11 months, 3 weeks ago

Since it's not given whether air drag is present or not, we assume that it's absent. The velocity components along the horizontal direction (assumed the x x -axis) and the vertical direction (assumed the y y -axis) are v x = v 0 cos ⁡ θ , v y = v 0 sin ⁡ θ − g t v_x=v_0\cos \theta, v_y=v_0\sin \theta -gt , t f = 2 v 0 sin ⁡ θ g t_f=\dfrac {2v_0\sin \theta}{g}

⟹ T ( t ) = 1 2 m ( v x 2 + v y 2 ) \implies T(t)=\dfrac {1}{2}m(v_x^2+v_y^2)

= 1 2 m v 0 2 − m v 0 g t sin ⁡ θ + 1 2 m g 2 t 2 =\dfrac {1}{2}mv_0^2-mv_0gt\sin \theta +\dfrac {1}{2}mg^2t^2

⟹ ∫ 0 t f T ( t ) d t = m v 0 3 sin ⁡ θ g ( 1 − 2 3 sin ⁡ 2 θ ) \implies \displaystyle \int_0^{t_f} T(t)dt=\dfrac {mv_0^3\sin \theta }{g}(1-\frac{2}{3}\sin^2 \theta)

The y y -coordinate of the projectile is v 0 t sin ⁡ θ − 1 2 g t 2 v_0t\sin \theta -\dfrac {1}{2}gt^2

⟹ V ( t ) = m v 0 g t sin ⁡ θ − 1 2 m g 2 t 2 \implies V(t)=mv_0gt\sin \theta-\dfrac {1}{2}mg^2t^2

⟹ ∫ 0 t f V ( t ) d t \implies \displaystyle \int_0^{t_f} V(t)dt

= 2 m v 0 3 sin ⁡ 3 θ 3 g =\dfrac {2mv_0^3\sin^3 \theta}{3g}

So, the ratio is η = 3 2 sin ⁡ 2 θ − 1 \eta=\dfrac {3}{2\sin^2 \theta}-1

For η = 1 , sin ⁡ 2 θ = 3 4 \eta=1, \sin^2 \theta=\dfrac {3}{4}

Since θ \theta is acute, sin ⁡ θ = 3 2 ⟹ θ = 60 ° \sin \theta=\dfrac {\sqrt 3}{2}\implies \theta=\boxed {60\degree} .

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