Determining angle and ratio in 2d motion

1. Sep 26, 2006

parwana

When baseball outfielders throw the ball, they usually allow it to take one bounce on the theory that the ball arrives sooner this way. Suppose that after the bounce the ball rebounds at the same angle as it had when released (Fig. P3.70) but loses half its speed.

(a) Assuming the ball is always thrown with the same initial speed, at what angle should the ball be thrown in order to go the same distance D with one bounce (blue path) as one thrown upward at = 49.6° with no bounce (green path)?
°

(b) Determine the ratio of the times for the one-bounce and no-bounce throws.
t1b / t0b =

2. Sep 27, 2006

andrevdh

(a) Use the fact that the range of the green projectile is equal to the range of the two blue projectiles. That is

$$D_o = D_1 + D_2$$

(b) This one gets messy, but it is doable. I used the x-component of the projectile's (initial) speed $v_o$ to do the calculations. Set up the times equations, $t_o,\ t_1\ ,t_2$, using the ranges of the projectiles. That is in terms of the D's. Then to get rid of the D's in your equations use the range formula again. You need to determine the ratio

$$\frac{t_1 + t_2}{t_o}$$

3. Sep 27, 2006

parwana

4. Sep 27, 2006

$$Range = \frac{v_{0}^{2}\sin2\phi}{g}$$ for the green path, so solve for $$\theta$$ You know for the blue path the range is $$\frac{v_{0}^{2}\sin 99.2}{g}$$ So $$\frac{v_{0}^{2}\sin2\theta}{g} + \frac{\frac{v_{0}}{4}^{2}\sin2\theta}{g} = \frac{v_{0}^{2}\sin 99.2}{g}$$ I got $$\theta \doteq 28.53$$

Last edited: Sep 27, 2006
5. Sep 27, 2006

Staff: Mentor

BTW, for extra credit you can point out that the premise of the question is incorrect. The person who wrote the question obviously never played baseball at any coached level. Outfielders throw home plate liners low with one bounce near the pitcher's mound so that infielders can cut it off. They can cut it off if the throw is fading and hit home with their own rope, or they can cut it off to nail other base runners if the play at home looks iffy.

6. Sep 27, 2006

parwana

courtigrad I am still not getting it right, thanks though

7. Sep 27, 2006

8. Sep 27, 2006

parwana

20.4 isnt right

9. Sep 27, 2006

what about $$28.53$$? it was supposed to be $$\frac{v_{0}^{2}}{4}$$

10. Sep 27, 2006

parwana

that was wrong too, I figured it out though, it was 26.5

11. Sep 27, 2006

parwana

Now I need help with part b

12. Sep 28, 2006

andrevdh

Calculate the times with the x-velocity component for each part of the motion:

$$t = \frac{v_x}{D_i}$$