An Uncommon Calculation

Classical Mechanics Level 3

A projectile is launched from level ground with speed v 0 v_0 at an angle θ \theta with respect to ground. The gravitational acceleration is g g .

Let t = 0 t = 0 be the launch instant, and let t f t_f be the time at which the projectile lands. If [ x ( t ) , y ( t ) ] [x(t),y(t)] are the coordinates of the projectile, calculate the following integral:

I = 1 0 − 6 ∫ 0 t f [ x 2 ( t ) + y 2 ( t ) ] d t \large{I = 10^{-6} \int_0^{t_f} \Big[x^2 (t) + y^2 (t) \Big] \, dt}

Details and Assumptions:
1) v 0 = 100 m/s v_0 = 100 \, \text{m/s}
2) g = 10 m/s 2 g = 10 \, \text{m/s}^2
3) θ = π 3 rad \theta = \frac{\pi}{3} \, \text{rad}
4) In the equations above, the brackets function like parentheses


The answer is 5.629.

Steven Chase by Steven Chase , 37, USA

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1 solution

Max Yuen
Jun 8, 2019

Uncommon Calculation meets uncommon solution:

Let's go unitless, so as x x goes from 0 to R R , and y y goes from 0 to h h to 0 again, it takes t t from 0 to t f t_f , and as a result we can have the following substitutions:

u = x R \large u = \frac{x}{R}

v = y R \large v = \frac{y}{R}

τ = t t f \large \tau = \frac{t}{t_f}

Where R = v 0 2 sin ⁡ 2 θ / g = 500 3 R = v_0^2\sin{2\theta}/g = 500\sqrt{3} and t f = 2 v 0 sin ⁡ θ / g = 10 3 t_f = 2v_0\sin{\theta}/g = 10\sqrt{3} .

The projectile trajectory is now

u = τ \large u = \tau

v = tan ⁡ θ ( 1 − u ) u = 3 ( 1 − u ) u \large v = \tan{\theta}(1-u)u=\sqrt{3}(1-u)u

Thus, the integral asked for is reduced to

I = 1 0 − 6 ⋅ R 2 t f ∫ 0 1 u 2 + v 2 d τ = 1 0 − 6 ⋅ R 2 t f ∫ 0 1 u 2 + 3 ( 1 − u ) 2 u 2 d u \large I = 10^{-6}\cdot R^2t_f\int_0^1{u^2+v^2}d\tau =10^{-6}\cdot R^2t_f\int_0^1{u^2+3(1-u)^2u^2}du

The integral on the right is easily done with the help of ∫ 0 1 u n d u = 1 n + 1 \int_0^1{u^n}du = \frac{1}{n+1}

Thus, I = 1 0 − 6 ⋅ ( 500 3 ) 2 ⋅ 10 3 ⋅ ( 3 ( 1 / 3 − 2 / 4 + 1 / 5 ) + 1 / 3 ) = 5.6292. \large I = 10^{-6}\cdot (500\sqrt{3})^2 \cdot 10\sqrt{3} \cdot \left(3(1/3-2/4+1/5)+1/3\right) = 5.6292.

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