Jason Kendall throws a baseball with a horizontal component of velocity 25 m/s. It takes 3.0 s to come back to its original height. Calculate its horizontal range, its initial vertical component of velocity and its initial angle of projection.

Horizontal Range: 75 m

Initial Vertical Component of Velocity: 0 m/s
Initial Angle of Projection: 0°

To find the horizontal range, initial vertical component of velocity, and initial angle of projection, we can solve the problem using the following steps:

Step 1: Find the time taken to reach the maximum height (t_max).
Step 2: Determine the maximum height reached (h_max).
Step 3: Calculate the initial vertical component of velocity (v_y0).
Step 4: Find the horizontal range (R).
Step 5: Calculate the initial angle of projection (θ).

Let's go through these steps one by one:

Step 1: Find the time taken to reach the maximum height (t_max).
Since the vertical motion follows a parabolic trajectory, the time taken to reach the maximum height is equal to half the total time of flight. Given that it takes 3.0 seconds to come back to its original height, the time taken to reach the maximum height is:
t_max = 3.0 s / 2
t_max = 1.5 s

Step 2: Determine the maximum height reached (h_max).
Using the equation of motion for vertical motion, we can find the maximum height using the formula:
h_max = v_y0 * t_max - (1/2) * g * t_max^2
where v_y0 is the initial vertical component of velocity and g is the acceleration due to gravity (g = 9.8 m/s^2).
Since the ball starts and ends at the same height, the maximum height reached is zero:
0 = v_y0 * 1.5 s - (1/2) * (9.8 m/s^2) * (1.5 s)^2

Solving the equation, we get:
v_y0 * 1.5 s = (1/2) * (9.8 m/s^2) * (1.5 s)^2
v_y0 = (1/2) * (9.8 m/s^2) * (1.5 s)
v_y0 = 7.35 m/s

Step 4: Find the horizontal range (R).
The horizontal range can be calculated using the equation:
R = v_x * t_total
where v_x is the horizontal component of velocity and t_total is the total time of flight.
Given that the horizontal component of velocity is 25 m/s and the total time of flight is 3.0 seconds, we have:
R = 25 m/s * 3.0 s
R = 75 m

Step 5: Calculate the initial angle of projection (θ).
The initial angle of projection can be found using the relationship between the horizontal and vertical components of velocity:
tan(θ) = v_y0 / v_x
Using the given values, we can calculate:
tan(θ) = 7.35 m/s / 25 m/s

Taking the inverse tangent (arctan) of both sides, we get:
θ = arctan(7.35 m/s / 25 m/s)

Using a calculator to find the inverse tangent, we get:
θ ≈ 16.95°

Therefore, the horizontal range is 75 meters, the initial vertical component of velocity is 7.35 m/s, and the initial angle of projection is approximately 16.95 degrees.

To find the answers to these questions, we need to use the kinematic equations for projectile motion.

Let's break down the given information:

1. Horizontal component of velocity = 25 m/s
2. Time of flight (time taken to reach the same height) = 3.0 s

Now, based on these details, let's find the answers step by step:

1. Horizontal range:
The horizontal range is the total distance traveled by the projectile in the horizontal direction when it returns to the same height. The formula for horizontal range is:
Range = horizontal velocity × time of flight

Given:
Horizontal component of velocity (Vx) = 25 m/s
Time of flight (t) = 3.0 s

Using the formula, plug in the values:
Range = Vx × t
Range = 25 m/s × 3.0 s
Range = 75 m

So, the horizontal range is 75 meters.

2. Initial vertical component of velocity:
To find the initial vertical component of velocity, we know that the time taken for the projectile to reach the same height is equal to the time taken for it to reach the highest point. The vertical motion can be analyzed using the formula:
Vy = Voy - g × t
where:
Vy = vertical component of velocity at any time
Voy = initial vertical component of velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

Given:
Time of flight (t) = 3.0 s
Acceleration due to gravity (g) = 9.8 m/s^2

Using the formula, plug in the values to find Voy:
0 = Voy - (9.8 m/s^2) × (3.0 s)
Voy = 29.4 m/s

Therefore, the initial vertical component of velocity is 29.4 m/s.

3. Initial angle of projection:
To find the initial angle of projection, we need to use the trigonometric relationship between the horizontal and vertical components of velocity. The angle of projection (θ) can be found using the following formula:
θ = tan^(-1)(Voy / Vx)

Given:
Voy = 29.4 m/s
Vx = 25 m/s

Using the formula, plug in the values:
θ = tan^(-1)(29.4 m/s / 25 m/s)
θ = 50.5° (rounded to one decimal place)

Therefore, the initial angle of projection is approximately 50.5°.

To summarize:
- The horizontal range is 75 meters.
- The initial vertical component of velocity is 29.4 m/s.
- The initial angle of projection is approximately 50.5°.