A baseball is thrown vertically into the air with a speed of 24.7 m/s. How high does it go? How long does the round trip up and down require?

Answer 1

First let's look at this scenario. We can treat the baseball as essentially a point mass so we can apply the equations of motion. Since it is projected vertically, we do not need to worry about any horizontal motion. The force this baseball is subjected to is the force due to gravity, which is essentially constant at the surface of the earth. This will accelerate the ball pulling it downwards as it goes up and as it comes down after it reaches a maximum height. We will ignore air resistance since it was not mentioned explicitly.

Let's call the initial velocity the baseball has u = 24.7 m/s

Let's call the acceleration due to gravity g = -10 m/s² (note this is approximate, and the negative sign indicates the acceleration is downwards, since we are defining upwards as positive)

If we define t = time to reach the peak height, then in this case y = peak height. For ease we can define the height at which the ball was launched as y₀ = 0.

v = velocity at the maximum height the ball reaches. Note that v = 0 here, since the ball comes to a momentary stop at the peak height before moving back down.

Let's look at the equations of motion and see which ones we will use:

1. v = u + a*t
2. y - y₀ = u*t + (1/2)*a*t²
3. v² = u² + 2*a*(y - y₀)²

Since we know v = 0, u = 24.7 m/s & a = -10 m/s², we can use the first equation to find the time to reach the maximum height. Let's rearrange it to make 't' the subject of the equation:

t = (v - u)/a = (0-24.7)/-10 = 2.47 seconds

Since we want the time to go up and come back down, we can double this time to get a final answer of 4.94 seconds.

To find the max height 'y', we can use the second equation (note the time to reach the max height is 2.47 s):

y - y₀ = u*t + (1/2)*a*t²

y - 0 = 24.7 * 2.47 + (1/2)* (-10)* (2.47)²

y = 30.5 m