When baseball players throw the ball in from the outfield, they sometimes allow it to take one bounce before it reaches the infield on the theory that the ball arrives sooner that way. Suppose the angle at which a bounced ball leaves the ground is the same as the angle at which the outfielder threw it as shown in the figure, but the ball’s speed after the bounce is ½ of what it was before the bounce. (a) Assume the ball is always thrown with the same initial speed. At what angle θ should the fielder throw the ball to make it go the same distance D with one bounce as a ball thrown upward at 45.0o with no bounce? (b) Determine the ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw (t of bounced ball / t of 45o ball)
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(a) To solve this problem, we can use the principles of projectile motion.
Let's first consider the ball thrown upward at 45.0o with no bounce. In this case, the initial speed of the ball is the same throughout its trajectory.
Given that the distance traveled by this ball is D, we can use the equations of projectile motion to find the initial speed and angle of launch for the ball thrown with one bounce.
The horizontal component of velocity remains the same before and after the bounce. Therefore, the horizontal component of velocity of the ball thrown with one bounce is the same as the horizontal component of velocity for the ball thrown upward at 45.0o.
Let's denote the initial speed of the ball thrown with one bounce as V before the bounce, and V/2 after the bounce.
Using the equation for horizontal velocity, we have:
V * cos(θ) = (V/2) * cos(45.0o)
Simplifying this equation, we find:
cos(θ) = cos(45.0o) / 2
Taking the inverse cosine of both sides, we can solve for the angle θ:
θ = cos^(-1)(cos(45.0o) / 2)
(b) To determine the ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw (t of bounced ball / t of 45o ball), we can compare the vertical components of their velocities.
For the ball thrown upward at 45.0o, we can determine the flight time using the equation:
t = 2 * (V * sin(45.0o) / g)
where g is the acceleration due to gravity.
For the ball thrown with one bounce, we can determine the time interval until the ball reaches the highest point before falling due to gravity. This time interval is half of the flight time because the speed after the bounce is half the initial speed.
Therefore, the ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw is 1/2.
Keep in mind that these calculations assume ideal conditions and neglect any other factors such as air resistance.
To solve both parts of the problem, we can use the concept of projectile motion. We'll break it down step-by-step:
(a) To find the angle θ at which the fielder should throw the ball to make it go the same distance D with one bounce as a ball thrown upward at 45.0° with no bounce, we need to find the initial speed of the ball after the bounce.
Let's assume the initial speed of the ball is v. After the bounce, the ball's speed is ½v, and the angle at which it leaves the ground is the same as the angle at which it was thrown.
For the ball thrown upward at 45.0° with no bounce, the range D is given by:
D = (v² / g) * sin(2θ)
Since we want the same range D for the one-bounce throw, the range would be given by:
D = (½v² / g) * sin(2θ)
To find the angle θ, we can equate the two equations:
(v² / g) * sin(2θ) = (½v² / g) * sin(2θ)
Canceling out common terms, we get:
sin(2θ) = ½ sin(2θ)
Dividing both sides by sin(2θ), we have:
1 = ½
This equation is not possible because it doesn't have a solution. Therefore, there is no angle at which the fielder can throw the ball to make it go the same distance D with one bounce as a ball thrown upward at 45.0° with no bounce.
(b) The time interval for the one-bounce throw can be found by adding the time it takes for the ball to reach the maximum height (t₁) and the time it takes for the ball to come back down (t₂).
For the ball thrown upward at 45.0° with no bounce, the total flight time is given by:
t = 2 * t₁
The time it takes for the ball to reach the maximum height can be found using the equation for vertical motion:
t₁ = (v * sin(45°)) / g
Let's assume the time it takes for the ball to come back down after the bounce is t.
The ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw can be calculated as:
(t of bounced ball / t of 45° ball) = (t₁ + t₂) / (2 * t₁)
Since the speed of the ball after the bounce is ½v, we can use this to find the time it takes for the ball to come back down:
t₂ = (½v * sin(45°)) / g
Substituting the values for t₁ and t₂, we have:
(t of bounced ball / t of 45° ball) = [(v * sin(45°) / g) + (½v * sin(45°) / g)] / (2 * (v * sin(45°) / g))
Simplifying the equation further, we get:
(t of bounced ball / t of 45° ball) = 1 + ½ = 1.5
Therefore, the ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw is 1.5.