A baseball is hit nearly straight up into the air with a speed of 24 m/s.

(a) How high does it go?
(b) How long is it in the air?

(a) How high does it go?

Well, hopefully it doesn't go high enough to join a traveling circus as an aspiring acrobat. But if we put that aside, we can use some physics to find out. When the baseball reaches its highest point, its velocity becomes zero. So, we can use the equation:

v^2 = u^2 + 2as,

where v is the final velocity (zero in this case), u is the initial velocity (24 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (the height we're looking for).

Plugging in the values, we get:

0 = (24)^2 + 2(-9.8)s,

0 = 576 - 19.6s,

19.6s = 576,

s = 576/19.6 ≈ 29.39 meters.

So, the baseball goes approximately 29.39 meters high. Just don't challenge it to a high jump competition!

(b) How long is it in the air?
Ah, the majestic flight of the baseball. To calculate the time of flight, we can use the equation:

v = u + at,

where v is the final velocity (zero again), u is the initial velocity (24 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight.

Substituting the values, we get:

0 = 24 - 9.8t,

9.8t = 24,

t = 24/9.8 ≈ 2.45 seconds.

So, the baseball is in the air for about 2.45 seconds. Enough time to tweet a quick joke before it comes back down!

To find the answers to these questions, we can use the laws of motion, specifically, the equations of motion for a vertically launched projectile.

(a) To determine how high the baseball goes, we need to calculate the maximum height (vertical displacement) reached by the ball.

We can use the equation for vertical displacement, which is given by:

Δy = v₀y * t + (1/2) * g * t²

Where:
Δy = vertical displacement or height
v₀y = initial vertical velocity
t = time of flight
g = acceleration due to gravity (-9.8 m/s²)

In this case, the initial vertical velocity, v₀y, is the velocity when the ball is hit. Since the ball is hit straight up, the initial vertical velocity is equal to the magnitude of the initial velocity, v₀, which is 24 m/s.

Substituting the values into the equation, we have:

Δy = 24 * t + (1/2) * (-9.8) * t²

To find the maximum height, we know that at the maximum height, the vertical velocity becomes zero. Therefore, v_fy = 0.

v_fy = v₀y + g * t

0 = 24 - 9.8 * t

Rearranging the equation, we get:

t = 24 / 9.8 ≈ 2.45 seconds

Now, we can substitute the value of t back into the equation for Δy to find the maximum height:

Δy = 24 * 2.45 + (1/2) * (-9.8) * (2.45)²

Calculating this expression will give us the answer to part (a), which is the height the ball reaches.

(b) To find the time of flight or how long the ball remains in the air, we can use the same equation:

v_fy = v₀y + g * t

Since v_fy = 0 at the highest point, we can solve for t:

0 = 24 - 9.8 * t

Simplifying the equation, we get:

t = 24 / 9.8 ≈ 2.45 seconds

Thus, the ball is in the air for approximately 2.45 seconds.

Now, you can substitute the values into the equations and calculate the answers. Make sure to double-check your calculations for accuracy.