Rate of Change of Angle of Projectile

[LandmarkCharlie]

I am trying to figure out this problem:

A ball is thrown at a certain angle of elevation from the horizontal (m). Determine the rate of change of change of m (the angle between the direction the ball is moving and horizontal) when the ball is at its highest point.

I have done the following:
(Note: m = angle measure, a = acceleration due to gravity, v_yo=Initial vertical velocity, v_x=horizontal velocity, h_o=initial height)

h(t) = -.5at² + v_yot + h_o

tan(m) = (vert. velocity)/(horiz. velocity)

tan(m) = (-at + v_yo)/v_x

tan(m) = -at/v_x+ v_yo/v_x

d(tan(m))/dt = -a/v_x

(dm/dt)*sec²(m) = -a/v_x

dm/dt = (-a/v_x)*cos²m

When the ball is at the highest point, the vertical velocity is 0. Therefore m=0.

dm/dt = (-a/v_x)*cos²0

dm/dt = -a/v_x
---------------------
This doesn't makes sense to me. If a = 9.8m/s² and v_x = 9.8, then dm/dt = -1. But what are the units? -1 radian per second is a very different speed than -1 degree per second, and both cos(0 degrees) and cos(0 radians) are equal to 1.

 

[Ishuda]

I am trying to figure out this problem:

A ball is thrown at a certain angle of elevation from the horizontal (m). Determine the rate of change of change of m (the angle between the direction the ball is moving and horizontal) when the ball is at its highest point.

I have done the following:
(Note: m = angle measure, a = acceleration due to gravity, v_yo=Initial vertical velocity, v_x=horizontal velocity, h_o=initial height)

h(t) = -.5at² + v_yot + h_o

tan(m) = (vert. velocity)/(horiz. velocity)

tan(m) = (-at + v_yo)/v_x

tan(m) = -at/v_x+ v_yo/v_x

d(tan(m))/dt = -a/v_x

(dm/dt)*sec²(m) = -a/v_x

dm/dt = (-a/v_x)*cos²m

When the ball is at the highest point, the vertical velocity is 0. Therefore m=0.

dm/dt = (-a/v_x)*cos²0

dm/dt = -a/v_x
---------------------
This doesn't makes sense to me. If a = 9.8m/s² and v_x = 9.8, then dm/dt = -1. But what are the units? -1 radian per second is a very different speed than -1 degree per second, and both cos(0 degrees) and cos(0 radians) are equal to 1.

First, we note that, as usual, we are assuming that the x component of the velocity is constant, i.e., v_x=v_x0 is a constant (otherwise you need the product rule/ quotient rule).

What's the problem 0 radians per second is the same rate of change as 0 degrees per second

To be sirius (since that answer may have put me in the dog house), about your question on units: The difference between using radians and using degrees changes the equations. For example, suppose we were using m in degrees. The equations are not as you have them. They would actually be, for example,
tan(\(\displaystyle \frac{\pi}{180}\)*m) = (vert. velocity)/(horiz. velocity)
and the general answer would change so that
dm/dt = \(\displaystyle \frac{180}{\pi}\)*[(-a/v_x)*cos²(\(\displaystyle \frac{\pi}{180}\)*m)]

 

[LandmarkCharlie]

First, we note that, as usual, we are assuming that the x component of the velocity is constant, i.e., v_x=v_x0 is a constant (otherwise you need the product rule/ quotient rule).

What's the problem 0 radians per second is the same rate of change as 0 degrees per second

To be sirius (since that answer may have put me in the dog house), about your question on units: The difference between using radians and using degrees changes the equations. For example, suppose we were using m in degrees. The equations are not as you have them. They would actually be, for example,
tan(\(\displaystyle \frac{\pi}{180}\)*m) = (vert. velocity)/(horiz. velocity)
and the general answer would change so that
dm/dt = \(\displaystyle \frac{180}{\pi}\)*[(-a/v_x)*cos²(\(\displaystyle \frac{\pi}{180}\)*m)]

Ishuda,

Thank you for your reply! I don't really understand your reply, though. Why is radians the default unit for the angle? Given the vertical and horizontal velocity, I could calculate m in radians or degrees. In other words, why isn't the default unit degrees, and then my equation would have to change to tan(180/pi*m) = (vert. velocity)/(horiz. velocity) when m was in radians?

Furthermore, do you see any mistakes in my calculations?

Thanks again for your reply. Hopefully you will write again!

Charlie

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[arup2196]

I am trying to figure out this problem:

A ball is thrown at a certain angle of elevation from the horizontal (m). Determine the rate of change of change of m (the angle between the direction the ball is moving and horizontal) when the ball is at its highest point.

I have done the following:
(Note: m = angle measure, a = acceleration due to gravity, v_yo=Initial vertical velocity, v_x=horizontal velocity, h_o=initial height)

h(t) = -.5at² + v_yot + h_o

tan(m) = (vert. velocity)/(horiz. velocity)

tan(m) = (-at + v_yo)/v_x

tan(m) = -at/v_x+ v_yo/v_x

d(tan(m))/dt = -a/v_x

(dm/dt)*sec²(m) = -a/v_x

dm/dt = (-a/v_x)*cos²m

When the ball is at the highest point, the vertical velocity is 0. Therefore m=0.

dm/dt = (-a/v_x)*cos²0

dm/dt = -a/v_x
---------------------
This doesn't makes sense to me. If a = 9.8m/s² and v_x = 9.8, then dm/dt = -1. But what are the units? -1 radian per second is a very different speed than -1 degree per second, and both cos(0 degrees) and cos(0 radians) are equal to 1.

Unit would be radian per second. As you apply d(tan(m))/dt = (dm/dt)*sec²(m)
All this calculus formulas depend upon the fact that limiting value of sinx/x approaches to 1 as x approaches to 0. And here x must be in radian. Otherwise sinx/x does not approaches to 1 as x approaches to 0.

 

[Deleted member 4993]

Unit would be radian per second. As you apply d(tan(m))/dt = (dm/dt)*sec²(m)
All this calculus formulas depend upon the fact that limiting value of sinx/x approaches to 1 as x approaches to 0. And here x must be in radian. Otherwise sinx/x does not approaches to 1 as x approaches to 0.

The original post is 4 yrs. old. Poster probably has become a rocket scientist by now!!

 

[Steven G]

The original post is 4 yrs. old. Poster probably has become a rocket scientist by now!!

Not everyone can become a rocket scientist like you.