Solving The Flight of a Ball: Examining Velocity & Time

[ForceBoy]

Homework Statement

A ball is launched at 10 meters per second at an angle of 45 degrees above the horizontal.
What was the duration of the ball's flight?
What was it's highest point? When?
What was the total distance traveled?
Assume air resistance is negligible.
Upwards and rightwards are positive.

Homework Equations

y = v_yi+½(at²)
v_f = v_i+at
x = vt

The Attempt at a Solution

I start by breaking the velocity vector into its rectangular components:
v_y = 10⋅sin(45°) = 5√2 meters per second
We know that sin(45°) = cos(45°), therefore v_x = v_y

Now that I know the velocities, using Earth's acceleration due to gravity I can solve for airtime using:
v_f = v_i+ at
(I am using vertical velocity)
I set the final velocity to zero and plug the other known values.
0 = 5√2 - 9.8(t)
-5√2 = -9.8(t)
t = -5√2/-9.8
t ≈ 0.72 seconds
This is the time it takes the ball to reach a vertical velocity 0. I know this will happen in the middle of the ball's flight. Therefore, to find the total time I just double this time.

v_total ≈ 1.44 seconds.

Now that I know the time, velocities, and acceleration, I can solve for everything else.

x = v_xt
x= 5√2(1.44)
x≈10.18 meters traveled

When solving for the highest, I already know when it will happen so:
y = v_yi+½(at²)
y = 5√2(0.72) -½(9.8)(0.72)²
y ≈ 2.55 meters above the horizontal

So to answer the questions:
The ball will travel a total of about 10.18 meters in 1.44 second. It will hit it's highest point after 0.72 seconds at an altitude of 2.55 meters above the horizontal.

My question:

I have learned that when solving for air time to use the third kinematic equation. This works because when the ball flies, it traces a parabola; the third kinematic equation is a quadratic equation. This being said, I wondered if I could solve for time as I did. I would believe so because of the rectangular components: They are orthogonal acting at the same time, you could think of the ball moving in two different straight lines at the same time (vertically and horizontally). The ball travels linearly in the vertical direction (as it does in the horizontal) and the first kinematic equation is linear. Therefore it works. Or at least it appears this way to me. Am I overlooking something?

Another question, I do not have good enough intuition about speed in meters per seconds to determine whether my answers are reasonable. I would appreciate it if someone could give me an example of an object the travels at 10 m/s .

 

[kuruman]

Hi ForceBoy and welcome to PF.

Am I overlooking something?

You have not overlooked anything and I should add that you have a good grasp of this subject.

I would appreciate it if someone could give me an example of an object the travels at 10 m/s .

The world record for the 100 m dash is a bit under 10 seconds. This means that a human running at top speed for a short spurt has an average speed of about 10 m/s. Also, a good number to remember is that 20 m/s is 45 miles per hour.

 

[ForceBoy]

Thank you very much!