Projectile Motion, requiring variable elimination

[Lukapaka]

Homework Statement

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 but loses half its speed.

1) Assuming the ball is always thrown with the same initial speed, at what angle θ should the ball be thrown in order to go the same distance D with one bounce (blue path) as one thrown upward at α = 49.8° with no bounce (green path)?

2) Determine the ratio of the times for the one bounce and no bounce throws.

Homework Equations

1) d = v x t
2) Vh = V x cos(theta)

The Attempt at a Solution

d1 = distance of throw without bounce
d2 = distance of throw with one bounce
d2 = da + db = distance traveled before bounce + distance traveled after bounce

d1 = V x (cos(49.8)) x t
d2 = V x (cos(alpha)) x ta + (V/2)x(cos(alpha)) x tb
d1 = d2
V x (cos(49.8)) x t = V x (cos(alpha)) x ta + (V/2)x(cos(alpha)) x tb
V cancels and I'm left with:
cos(49.8) x t = (cos(alpha)) x ta + (1/2)x(cos(alpha)) x tb

I can't get the times to cancel out, and don't know any other way to approach this question. Help!

 

[v0id19]

this one problem can be broken down into two: a problem in the x direction (the one you're interested in), and a problem in the y direction.

time is the same for both problems (since they both are really one big problem, they start and end at the same time)

thus, find an equation involving time for the y-problem, solve for t and substitute it into your y-problem's equation

 

[Lukapaka]

1st throw
y = [(-9.8) x t^2] / 2 + V x sin (theta) x t
y = 0 at t = 0 and t = ?
theta = 49.8 deg
V x sin(49.8) x t - (4.9) x t^2 = 0
t[V x sin(49.8) - (4.9) x t] = 0
V x sin(49.8) - (4.9) x t = 0
V x sin(49.8) = (4.9) x t
t = V x sin(49.8) / (4.9)

X1 = Distance of 1st throw
X1 = V x cos (theta) x t
X1 = V x cos(49.8) x V x sin(49.8) / 4.9
X1 = V^2 (0.1006)

2nd throw
1st arch
y = [(-9.8) x t^2] / 2 + V x sin (theta) x t
y = 0 at t = 0 and t = ?
theta = @
V x sin(@) x t - (4.9) x t^2 = 0
t[V x sin(@) - (4.9) x t] = 0
V x sin(@) - (4.9) x t = 0
V x sin(@) = (4.9) x t
t = V x sin(@) / (4.9)

Xa = V x cos (theta) x t
Xa = V x cos (@) x V x sin(@) / (4.9)
Xa = V^2 x cos (@) x sin(@) / 4.9

2nd arch
y = [(-9.8) x t^2] / 2 + V x sin (theta) x t
y = 0 at t = 0 and t = ?
theta = @
(V/2) x sin(@) x t - (4.9) x t^2 = 0
t[(V/2) x sin(@) - (4.9) x t] = 0
(V/2) x sin(@) - (4.9) x t = 0
(V/2) x sin(@) = (4.9) x t
t = V x sin(@) / (9.8)

Xb = V x cos (theta) x t
Xb = V x cos (@) x V x sin(@) / (9.8)
Xb = V^2 x cos (@) x sin(@) / 9.8

X2 = Xa + Xb = V^2 x cos (@) x sin(@) / 4.9 + V^2 x cos (@) x sin(@) / 9.8
X2 = 3 x V^2 x cos(@)sin(@) / 9.8
X2 = (.30612) x V^2 x cos(@)sin(@)

X1 = X2
V^2 (0.1006) = (.30612) x V^2 x cos(@)sin(@)
.1006 = .30612 x cos(@)sin(@)
cos(@)sin(@) = 0.32866

I still can't figure this out.

 

[cryptlogic]

t-> time for no bounce D-> distance for no bounce
t1-> time for through b4 bounce d1-> distance for through b4 bounce
t2->time for after bounce d2-> distance for after bounce


V= D / t => t= D / V }
}
V= d1 / t1 => t1= d1 / V } ==> t / (t1 + t2) = D / (d1 + d2) --> ratio of times
}
V= d2 / t2 => t2= d2 / V }

 

[andrevdh]

Haven't you derived an equation for the range of a projectile?

 

[rl.bhat]

X1 = V x cos(49.8) x V x sin(49.8) / 4.9

It should be
X1 = V x cos(49.8) x V x sin(49.8) / 9.8
X2 = Xa + Xb = V^2 x cos (@) x sin(@) / 4.9 + V^2 x cos (@) x sin(@) / 9.8
It should be

X2 = Xa + Xb = V^2 x cos (@) x sin(@) /9.8 + V^2/4 x cos (@) x sin(@) / 9.8

 

[cryptlogic]

Haven't you derived an equation for the range of a projectile?

No, I do not think it is derived through that method...although it is still not entirely accurate for the above stated problem, but it is a start point to the solution.