Determining angle and ratio in 2d motion

[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 =

 

[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}

 

[parwana]

I don't understand, please help more

 

[courtrigrad]

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

 

[berkeman]

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.

 

[parwana]

courtigrad I am still not getting it right, thanks though

 

[courtrigrad]

what does the answer say?

 

[parwana]

20.4 isn't right

 

[courtrigrad]

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

 

[parwana]

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

 

[parwana]

Now I need help with part b

 

[andrevdh]

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

t = \frac{v_x}{D_i}