A baseball is thrown at an angle of 20◦ relative to the ground at a speed of 25.6 m/s. The ball is caught 42.9416 m from the thrower.

The acceleration due to gravity is 9.81 m/s2 .
How long is it in the air?

To find the time the baseball is in the air, we can use the equation of motion in the vertical direction:

y = y0 + v0y * t - (1/2) * g * t^2

where:
- y is the vertical displacement (which is 0 since the ball starts and ends at the same height)
- y0 is the initial vertical position (also 0)
- v0y is the initial vertical velocity
- g is the acceleration due to gravity
- t is the time the ball is in the air

Since the ball is thrown at an angle of 20° relative to the ground, the initial velocity can be split into its horizontal and vertical components:

v0x = v0 * cos(theta)
v0y = v0 * sin(theta)

where:
- v0 is the initial velocity of the ball
- theta is the launch angle (20°)

Plugging these values into the equation of motion, we have:

0 = 0 + v0y * t - (1/2) * g * t^2

Simplifying:

v0y * t = (1/2) * g * t^2

Dividing both sides by t:

v0y = (1/2) * g * t

Solving for t:

t = v0y / [(1/2) * g]

We can calculate v0y using the initial velocity and the launch angle:

v0y = v0 * sin(theta)

Substituting this into the equation for t:

t = (v0 * sin(theta)) / [(1/2) * g]

Let's calculate the time the baseball is in the air. Plugging in the values:

v0 = 25.6 m/s
theta = 20°
g = 9.81 m/s^2

t = (25.6 * sin(20°)) / [(1/2) * 9.81] = 1.768 seconds

Therefore, the baseball is in the air for approximately 1.768 seconds.

To find the time the baseball is in the air, we can use the horizontal motion and vertical motion separately.

First, let's consider the horizontal motion. The horizontal component of the initial velocity (25.6 m/s) remains constant throughout the flight, as no horizontal forces act on the baseball. Using this, we can calculate the time it takes for the ball to travel a horizontal distance of 42.9416 m.

The horizontal distance (x) can be calculated using the formula:
x = v * t
where x is the distance, v is the initial horizontal velocity, and t is the time.

Rearranging the formula, we can solve for time (t):
t = x / v

Plugging in the given values, we have:
t = 42.9416 m / 25.6 m/s

Calculating this gives us:
t ≈ 1.679 s

Therefore, the horizontal time of flight is approximately 1.679 seconds.

Now, let's consider the vertical motion. The projectile follows a parabolic path, and we'll use the vertical motion equation to find the time of flight.

The vertical distance (y) at any time (t) can be calculated using the formula:
y = v0 * t + (1/2) * a * t^2
where y is the distance, v0 is the initial vertical velocity, a is the acceleration due to gravity (-9.81 m/s^2), and t is the time.

The initial vertical velocity can be calculated using the initial velocity (25.6 m/s) and the launch angle (20 degrees). We can break down the initial velocity into its vertical and horizontal components using trigonometry.

The vertical component of the initial velocity (v0) can be calculated using the formula:
v0 = v * sin(angle)
where v is the initial velocity and angle is the launch angle.

Plugging in the given values, we have:
v0 = 25.6 m/s * sin(20°)

Calculating this gives us:
v0 ≈ 8.710 m/s

Now we can calculate the time of flight. The ball is launched vertically and returns to the same height, so the final vertical distance (y) is 0:

0 = v0 * t + (1/2) * a * t^2

Rearranging the equation, we have:
(1/2) * a * t^2 = -v0 * t

Plugging in the values, we get:
(1/2) * (-9.81 m/s^2) * t^2 = -(8.710 m/s) * t

Simplifying, we have:
-4.905 m/s^2 * t^2 = -8.710 m/s * t

Dividing both sides by t gives us:
-4.905 m/s^2 * t = -8.710 m/s

Solving for t gives us:
t = -8.710 m/s / -4.905 m/s^2

Calculating this gives us:
t ≈ 1.777 s

Therefore, the vertical time of flight is approximately 1.777 seconds.

Since we previously found that the horizontal time of flight is 1.679 seconds, we can conclude that the total time of flight is the longer of the two: approximately 1.777 seconds.

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