# Rate of change of hypotenuse

### Rate of change of hypotenuse

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In the problem, there is a projectile(1) moving at 5m/s in a line parallel to the x axis at y=10. Given that at time 0 projectile 1 is at (0,10), if a second projectile(2) were to be fired at 10m/s from (0,0) at an angle to intercept projectile 1, how can the distance traveled by projectile 2 be calculated at a given time (t)?
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### Re: Rate of change of hypotenuse

After t seconds projectile 1 is at (5t, 10) and projectile 2 is at (10t cos(a), 10t sin(a)) where a is the angle the trajectory of projectile 2 makes with the horizontal. They will collide at time T where 5T= 10T cos(a) and 10T sin(a)= 10. cos(a)= 5/10= 1/2 so a= 60 degrees. $sin(a)= sin(60)= \sqrt{3}/2$ so at time t, projectile 2 is at $(5t,\sqrt{3}t/2)$.

The distance projectile 2 will have traveled is $\sqrt{(5t)^2+ (\sqrt{3}t/2)^2}= \sqrt{25t^2+ 3t^2/4}= t\sqrt{103}/2$.
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