- #1
nineeyes
- 21
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Problem : A baseball player releases a ball with [tex]v_0= 100 ft/sec[/tex] at an angle of [tex]\theta= 30 degrees[/tex]. Determine the radius of curvature of the trajectory (a) just after release and (b) at the apex. For each case compute the time rate of change of the speed.
I really wasn't sure how to do the problem, so I sort of just guessed at a lot of things. So any help would be great. But this is what I did.
I assumed the only acceleration after the ball was thrown was due to gravity.
so I got
[tex] a_n=-32.2*cos(30)[/tex] [tex]a_t=-32.2*sin(30)[/tex]
then I used [tex] a_n=\frac{v^2}{\rho}[/tex] to solve for the curvature [tex]\rho[/tex] for when the ball was just released I got [tex]\rho = 358.6 ft[/tex] . Then I used [tex]\beta'=\frac{100}{\rho}[/tex] to solve for the angular rate.
I wasnt sure on how to solve (b) so I just assumed that the angle between [tex]\theta[/tex] and the x- axis would be 0 degrees since the apex is the highest point and I guessed that would be when the ball would stop moving up and start going down.
So I got
[tex]a_n = -32,2 ft/s^2[/tex]
then used
[tex]a_n=v*\beta'[/tex] to solve for v at the apex using the the angular rate i got earlier for beta'.
I got [tex]v=115.4ft/s[/tex]
then using the v and [tex]a_n[/tex] I solved for [tex]\rho[/tex] using the equation [tex] a_n=\frac{v^2}{\rho}[/tex]
I got [tex]\rho= 413 ft[/tex]
I'm definitely not confident with this answer, it's the first time I've done a problem like this, and I'm not sure if everything I did and assumed was legal.
Thanks in advance for any help.
I really wasn't sure how to do the problem, so I sort of just guessed at a lot of things. So any help would be great. But this is what I did.
I assumed the only acceleration after the ball was thrown was due to gravity.
so I got
[tex] a_n=-32.2*cos(30)[/tex] [tex]a_t=-32.2*sin(30)[/tex]
then I used [tex] a_n=\frac{v^2}{\rho}[/tex] to solve for the curvature [tex]\rho[/tex] for when the ball was just released I got [tex]\rho = 358.6 ft[/tex] . Then I used [tex]\beta'=\frac{100}{\rho}[/tex] to solve for the angular rate.
I wasnt sure on how to solve (b) so I just assumed that the angle between [tex]\theta[/tex] and the x- axis would be 0 degrees since the apex is the highest point and I guessed that would be when the ball would stop moving up and start going down.
So I got
[tex]a_n = -32,2 ft/s^2[/tex]
then used
[tex]a_n=v*\beta'[/tex] to solve for v at the apex using the the angular rate i got earlier for beta'.
I got [tex]v=115.4ft/s[/tex]
then using the v and [tex]a_n[/tex] I solved for [tex]\rho[/tex] using the equation [tex] a_n=\frac{v^2}{\rho}[/tex]
I got [tex]\rho= 413 ft[/tex]
I'm definitely not confident with this answer, it's the first time I've done a problem like this, and I'm not sure if everything I did and assumed was legal.
Thanks in advance for any help.