Eric C. answered 12/14/16
Tutor
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(117)
Engineer, Surfer Dude, Football Player, USC Alum, Math Aficionado
Hi Bailey.
After 2 seconds, the ball has a vertical altitude of:
y(2) = 200(2) - 40(2)2 = 400 - 160 = 240ft
The horizontal component of velocity is constant throughout the flight. After two seconds, the ball has traveled a horizontal distance of:
80(2) = 160ft
If we draw an angle to the horizontal from the start point to the ball, we can see that:
tan(Θ) = y/x
So
Θ = arctan(y/x)
You want to know how fast the angle is changing. In other words, you want dΘ/dt.
If we want to find out what dΘ/dt is, we can implicitly differentiate both sides of the equation.
d/dt(Θ = arctan(y/x))
dΘ/dt = 1/(1+(y/x)^2) * (x*dy/dt - y*dx/dt)/x2
Remember that the derivative of arctan(x) is 1/(1+x2). In this case, we don't have x, we have y/x, so we need to chain rule this quotient.
We can clean this expression up a little bit.
dΘ/dt = x2/(x2 + y2) * (x*dy/dt - y*dx/dt)/x2
dΘ/dt = (x*dy/dt - y*dx/dt)/(x2 + y2)
We know what y is (240 ft).
We know what x is (160 ft).
dx/dt is the rate of change of the horizontal component, which we know is a constant 80.
What about dy/dt?
y = 200t - 40t2
So
dy/dt = 200 - 80t
We want dy/dt when t = 2 seconds.
dy/dt(2) = 200 - 80*2 = 200 - 160 = 40
Plug in all these values to your dΘ/dt equation.
y = 240
x = 160
dx/dt = 80
dy/dt = 40
dΘ/dt = (x*dy/dt - y*dx/dt)/(x2 + y2)
= (160*40 - 240*80)/(1602 + 2402)
= -12800/83200
= -0.15385 rad/ sec
= -8.8 deg/ sec
Your angle is decreasing at a rate of 8.8 deg per second. Hope this helps (and is correct!).


Eric C.
tutor
Thanks Kenneth! I just hope it wasn't wrong! This is one of the few answers I'm mildly unsure of. You have any idea?
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12/14/16
Kenneth S.
12/14/16