# Projectile Motion, requiring variable elimination

• Lukapaka
In summary, the conversation discussed the problem of finding the angle and ratio of times for a baseball thrown with and without a bounce, with given initial conditions and equations. The solution involved breaking the problem into two separate problems in the x and y direction, finding an equation involving time for each, and then solving for the angle and ratio of times.
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!

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

Last edited:
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.

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 }

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

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

andrevdh said:
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.

## 1. What is projectile motion?

Projectile motion is the motion of an object in a curved path under the influence of gravity. It is a combination of horizontal and vertical motion, and is affected by the initial velocity, angle of launch, and gravity.

## 2. How do you calculate the maximum height of a projectile?

The maximum height of a projectile can be calculated using the equation h = (v02sin2θ)/2g, where h is the maximum height, v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

## 3. What is the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the equation R = v02sin2θ/g, where R is the range, v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

## 4. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the projectile and altering its trajectory. This is because air resistance creates a force opposite to the direction of motion, causing the projectile to lose speed and change direction.

## 5. Can you eliminate variables when calculating projectile motion?

Yes, variables can be eliminated when calculating projectile motion if certain conditions are met. For example, if the projectile is launched from ground level, the initial height can be assumed to be zero and eliminated from the equations. Additionally, if the launch angle is 45 degrees, the vertical and horizontal components of the initial velocity will be equal, allowing for the elimination of one of these variables.

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