A baseball player throwing the ball from the outfield usually allows it to take one bounce on the theory that the ball arrives sooner this way. Suppose that ball hits the ground at an angle θ and then bonuses but loses half of its speed.

A. Assuming that the ball is always at the same initial speed, at what angle θ should it be thrown in order to go the same distance D with one bounce as a ball thrown upward at 45° which reaches its target without bouncing

B. Determine the ration of the time for the one-bounce and the no-bounce throws.

To solve this problem, we need to consider the motion of the ball in both cases and use the principles of projectile motion. Let's break down the problem step by step:

A. To find the angle θ at which the ball should be thrown in the one-bounce scenario to cover the same distance D as the no-bounce throw, we can use the range formula for projectile motion:

Range (R) = (v^2 * sin(2θ)) / g

Here, v is the initial velocity of the ball, θ is the angle of projection, and g is the acceleration due to gravity.

For the no-bounce throw with an angle of 45°, the range can be expressed as:

Range (no-bounce) = (v^2 * sin(90)) / g
Range (no-bounce) = (v^2) / g

In the one-bounce scenario, the ball loses half of its speed after the bounce. So the speed of the ball after the bounce will be (v/2). The angle of projection after the bounce will be (180 - θ) due to the reflection of the ball.

The range for the one-bounce throw can be expressed as:

Range (one-bounce) = [(v/2)^2 * sin(2(180 - θ))] / g
Range (one-bounce) = [(v^2)/4 * sin(360 - 2θ)] / g
Range (one-bounce) = (v^2 * sin(2θ)) / (4g) ...[Using sin(θ) = sin(180 - θ)]

Since we want both throws to cover the same distance D, we equate the two ranges:

(v^2) / g = (v^2 * sin(2θ)) / (4g)

Simplifying the equation, we get:

v^2 = 4 * sin(2θ)
sin(2θ) = v^2 / 4
2θ = sin^(-1)(v^2 / 4)
θ = (sin^(-1)(v^2 / 4)) / 2

So, to throw the ball from the outfield with one bounce to cover the same distance as the no-bounce throw, the angle θ should be equal to (sin^(-1)(v^2 / 4)) / 2.

B. To determine the ratio of the time for the one-bounce and the no-bounce throws, we can use the time of flight formula:

Time of Flight = (2v * sin(θ)) / g

For the no-bounce throw, the time of flight can be expressed as:

Time of Flight (no-bounce) = (2v * sin(45°)) / g
Time of Flight (no-bounce) = (v * √2) / g

For the one-bounce throw, the time of flight can be divided into two parts: the time to reach the maximum height before the bounce and the time to come back down to the ground after the bounce.

Time of Flight (one-bounce) = [(2v * sin(θ)) / g] + [(2(v/2) * sin(180 - θ)) / g]
Time of Flight (one-bounce) = [(2v * sin(θ)) / g] + [(v * sin(180 - θ)) / g]
Time of Flight (one-bounce) = [(2v * sin(θ)) / g] + [(v * sin(θ)) / g]
Time of Flight (one-bounce) = [3v * sin(θ)] / g

The ratio of the time for the one-bounce and the no-bounce throws can be calculated as:

Ratio = [(3v * sin(θ)) / g] / [(v * √2) / g]
Ratio = [3 * sin(θ)] / √2

So, the ratio of the time for the one-bounce and no-bounce throws is [3 * sin(θ)] / √2.