A baseball is thrown at an angle of 25 ◦ relative to the ground at a speed of 27.6 m/s. The ball is caught 59.4844 m from the thrower. The acceleration due to gravity is 9.81 m/s 2 . How long is it in the air? Answer in units of s

the horizontal component of the velocity is

27.6 cos 25° = 25.01409 m/s

It traveled for 59.4844/25.01409 = 2.378 sec

To find the time the baseball is in the air, we can break down the initial velocity of the baseball into its horizontal and vertical components.

Given:
Initial speed (magnitude) of the baseball, v₀ = 27.6 m/s
Launch angle, θ = 25°
Acceleration due to gravity, g = 9.81 m/s²
Distance from thrower to catcher, d = 59.4844 m

Step 1: Find the vertical component of the initial velocity
v₀y = v₀ * sin(θ)

Step 2: Calculate the time it takes for the ball to reach the highest point of its trajectory
The time taken to reach the maximum height is the time it takes for the initial vertical velocity to reach zero.
0 = v₀y - g * t_max
t_max = v₀y / g

Step 3: Calculate the time it takes for the ball to fall back to the ground
The time taken to fall back to the ground is twice the time taken to reach the maximum height.
t_total = 2 * t_max

Step 4: Find the total time the ball is in the air
The total time is the time it takes for the ball to reach the catcher.
t_total = d / (v₀ * cos(θ))

Now, let's calculate the time the ball is in the air.

v₀y = v₀ * sin(25°)
v₀y = 27.6 * sin(25°) ≈ 11.61 m/s

t_max = v₀y / g
t_max = 11.61 / 9.81 ≈ 1.18 s

t_total = 2 * t_max
t_total = 2 * 1.18 ≈ 2.36 s

Therefore, the ball is in the air for approximately 2.36 seconds.

To find the time the baseball is in the air, we can use the horizontal and vertical components of its velocity.

Given:
Initial speed of the baseball, v0 = 27.6 m/s
Throwing angle, θ = 25°
Horizontal distance, x = 59.4844 m
Acceleration due to gravity, g = 9.81 m/s^2

First, we need to find the initial vertical velocity, v0y, and the initial horizontal velocity, v0x.

We can use the trigonometric functions to break down the initial velocity into its vertical and horizontal components:

v0x = v0 * cos(θ)
v0y = v0 * sin(θ)

Substituting the given values into these equations, we have:

v0x = 27.6 m/s * cos(25°)
v0y = 27.6 m/s * sin(25°)

Next, we can calculate the time it takes for the ball to reach the horizontal distance x. The horizontal distance traveled can be found using the equation:

x = v0x * t

Rearranging the equation, we have:

t = x / v0x

Substituting the given values into this equation, we get:

t = (59.4844 m) / (27.6 m/s * cos(25°))

Finally, we can solve for t:

t ≈ 2.628 seconds

Therefore, the ball is in the air for approximately 2.628 seconds.