A particle is launched from the origin of the $xyz$ coordinate system at time $t = 0$ with speed $50 \text{m/s}$ . The ambient gravitational acceleration is $10 \text{m/s}^2$ in the negative $z$ direction.

Suppose the particle passes through point $(x,y,z) = (5,7,10)$ . There are two times at which this can occur. Give your answer as the ratio of the larger time value to the smaller time value.

The answer is 36.36.

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The equations of motion of the particle projected from the origin are:

$\ddot{x} = 0 \ ; \ \dot{x}(0) = v_{ox} \ ; \ x(0) = 0$ $\ddot{y} = 0 \ ; \ \dot{y}(0) = v_{oy} \ ; \ y(0) = 0$ $\ddot{z} = -10 \ ; \ \dot{z}(0) = v_{oz} \ ; \ z(0) = 0$

Solving by double integration and applying the initial conditions leads to the solution:

$x = v_{ox}t \ ; \ y = v_{oy}t \ ; \ z = v_{oz}t - 5t^2$

Therefore, when the particle crosses the point $(5,7,10)$ :

$5 = v_{ox}t \ ; \ 7 = v_{oy}t \ ; \ 10 = v_{oz}t - 5t^2$ $v_{ox} = \frac{5}{t} \ ; \ v_{oy} = \frac{7}{t} \ ; \ v_{oz} = \frac{10+5t^2}{t}$

Now, at time $t=0$ , the particle is projected with a speed of $50 \ \mathrm{m/s}$ . This means:

$v_{ox}^2 + v_{oy}^2 + v_{oz}^2 = 2500$

Replacing the unknown velocity components with the equations above lead to an equation in terms of the unknown time:

$\frac{25}{t^2} + \frac{49}{t^2} + \frac{(10+5t^2)^2}{t^2}=2500$

Simplifying:

$25t^4 - 2400t^2 + 174 = 0$

This is a quadratic in $t^2$ using which $t$ can be conveniently solved for. The final answer is $\approx\boxed{36.361}$