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 theta as it had when released, but loses half its speed.
Assuming the ball is alyways 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 (lower path) as one thrown upward at angle 47.8 with no bounce (upper path?)
the only hint my teacher gave me was: solve this problem symbolically before you plug in numbers.
i don't even know where to start.
Symbolically:
Throw a ball up vertically, it stays in the air a long time, but doesn't go far. The energy of the throw concentrated the speed vertically, where it did little horizontal movement.
So when one angles a ball upwards, it is in the air longer, but that means the horizontal velocity is lower.
To get to a distance D, horizontal velocity needs to be maximum. So ignorign friction, the ball thrown horizontally is the way to go, letting it bounce/roll (neglicting friction). But friction is real, so with each bounce one looses energy.
So the question is, loosing 3/4 energy on one bounce (velocity is 1/2 before bounce), is it still a better deal to let it bounce to get the ball there faster.
Look at time to plate, which is determined by horzontal velocity .
To solve this problem symbolically, let's break it down step-by-step:
Step 1: Set up the equations for the ball's motion in both scenarios.
In the upper path scenario (no bounce), we can use the equations of motion to describe the ball's trajectory. The horizontal distance traveled (D) can be expressed as:
D = V₀ * cosθ * t
where V₀ is the initial speed and θ is the angle at which the ball is thrown.
In the lower path scenario (one bounce), we need to consider the vertical and horizontal motion of the ball separately. The horizontal distance traveled before the bounce can be expressed as:
D/2 = (V₀/2) * cosθ * t₁
where t₁ is the time taken for the ball to reach the ground (since the ball will reach the same height as it started). Note that the initial speed has been halved (V₀/2) due to the rebound.
After the bounce, the ball will travel a distance of D/2 again, but at a different angle (θ') and a reduced speed (V₀/2). The horizontal distance traveled after the bounce can be expressed as:
D/2 = (V₀/2) * cosθ' * t₂
where t₂ is the time taken for the ball to reach the final destination.
Step 2: Find a relation between θ and θ'.
Since the ball rebounds at the same angle (θ') as it had when released, the vertical component of the velocity should also be the same before and after the bounce. Therefore, the vertical velocities just before and after the bounce should be equal.
Given that the ball loses half of its speed after the bounce, the vertical component of the velocity after the bounce is (V₀/2) * sinθ', and before the bounce, it is V₀ * sinθ.
Equating these two velocities, we have:
V₀ * sinθ = (V₀/2) * sinθ'
Simplifying this expression, we find:
2sinθ = sinθ'
Step 3: Express θ' in terms of θ.
Using the trigonometric identity sin(2θ) = 2sinθcosθ, we can rewrite the equation from Step 2 as:
sin(2θ) = sinθ'
Applying the inverse sine function to both sides of the equation, we get:
2θ = θ'
Step 4: Determine the value of θ that satisfies the condition.
Now let's plug the relationship we found in Step 3 (θ' = 2θ) into the equation for the horizontal distance traveled after the bounce (D/2 = (V₀/2) * cosθ' * t₂). We obtain:
D/2 = (V₀/2) * cos(2θ) * t₂
Since D and V₀ are constants, we don't need to consider them further. Dividing both sides of the equation by (V₀/2) and rearranging, we have:
cos(2θ) = (D/2) / (V₀/2) * t₂
Simplifying this expression, we find:
cos(2θ) = Dt₂ / V₀
Step 5: Determine the value of t₂.
To find the value of t₂, we use the equation of motion for the vertical motion during the bounce. The ball reaches the same height as it started, so its vertical displacement is zero:
0 = (V₀/2) * sinθ' * t₂ - (1/2) * g * (t₂^2)
where g is the acceleration due to gravity. Rearranging this equation, we obtain:
t₂ = 2(V₀ * sinθ') / g
Step 6: Substitute the value of t₂ in terms of θ' into Step 5.
Substituting the value of t₂ from Step 5 into the equation in Step 4, we get:
cos(2θ) = D * (2(V₀ * sinθ') / g) / V₀
Simplifying this expression, we find:
cos(2θ) = 2D * sinθ' / g
Since θ' = 2θ (from Step 3), we can rewrite the equation as:
cos(2θ) = 2D * sin(2θ) / g
Step 7: Solve for θ.
Rearranging and simplifying this expression further, we get:
cot(2θ) = 2D / g
Finally, solving for θ, we find:
2θ = cot^(-1)(2D/g)
Therefore, the angle at which the ball should be thrown to go the same distance with one bounce as thrown upward at angle 47.8° with no bounce is:
θ = cot^(-1)(2D/g) / 2
To solve this problem, let's break it down step by step:
Step 1: Understand the problem
The problem states that a baseball outfielder throws the ball with the same initial speed in two different scenarios. In the first scenario, the ball is thrown upward at an angle of 47.8 degrees with no bounce (upper path). In the second scenario, the outfielder allows the ball to bounce, and it rebounds at the same angle (theta) with half its initial speed. We need to determine at what angle (theta) the ball should be thrown in order to cover the same distance (D) as the first scenario.
Step 2: Establish variables and equations
Let's define some variables:
- v0: initial speed of the ball
- g: acceleration due to gravity (approximately 9.8 m/s^2)
- theta: angle at which the ball is thrown in the second scenario (with bounce)
The key equations we can use are:
- Horizontal distance (D) = v0 * (total time of flight)
- Vertical displacement = (v0 * t) * sin(angle) - (1/2) * g * t^2 (for the upper path with no bounce)
- Vertical displacement = (v0 * t) * sin(theta) - (1/2) * g * t^2 (for the lower path with one bounce)
Step 3: Solve the problem symbolically
We need to make the horizontal distances covered in both scenarios equal. Therefore, we set up the equation:
v0 * (total time of flight for the upper path) = v0 * (total time of flight for the lower path)
To calculate the time of flight, we can use the vertical displacements equations. Equate the vertical displacement for both scenarios, without substitution:
(v0 * t) * sin(47.8) - (1/2) * g * t^2 = (v0 * t) * sin(theta) - (1/2) * g * t^2
Simplify the equation by canceling out common terms:
(sin(47.8) * t) = (sin(theta) * t)
Since time (t) is non-zero, we can cancel it out from both sides:
sin(47.8) = sin(theta)
Step 4: Solve for theta
Now we need to find the angle (theta) that satisfies the equation sin(47.8) = sin(theta). Remember that the sine function is periodic with period 360 degrees (or 2π radians). Thus, angles that differ by multiples of 360 degrees (or 2π radians) will have the same sine value.
Therefore, to solve for theta, we can use the inverse sine function (also known as arcsine):
theta = arcsin(sin(47.8))
Use a calculator to find the arcsine of sin(47.8). Keep in mind that the arcsine function gives an angle in radians, so you may need to convert it to degrees if necessary.
Step 5: Plug in numbers if needed
The problem specifically asks you to solve symbolically first, so you don't need to plug in any numbers at this point. However, if you have specific values for D, v0, or g, you can substitute them into the equations to find the numerical answer.
Remember, if you substitute values, ensure consistent units of measurement (e.g., meters for D, m/s for v0, etc.).
I hope this explanation helps you understand the problem and how to approach it. Good luck with your calculations!