A ball is launched straight up into the air with a speed of 22 m/s. How long will it take to fall back to the ground? Round to the nearest 0.1s?

To solve this problem, we'll use the equations of motion under constant acceleration due to gravity. The ball is launched straight up, and the acceleration due to gravity will slow it down until it reaches its peak, then bring it back down.

Given:
- Initial velocity \( u = 22 \, \text{m/s} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) (downwards)
- Final velocity when the ball reaches its peak height \( v = 0 \, \text{m/s} \)

First, we calculate the time it takes to reach the peak height:

\[
v = u - g \times t
\]
\[
0 = 22 - 9.8 \times t
\]
\[
t = \frac{22}{9.8} \approx 2.24 \, \text{seconds}
\]

This is the time to reach the peak. Since the time to fall back down is the same, the total time for the entire trip (up and down) is:

\[
\text{Total time} = 2 \times 2.24 \approx 4.5 \, \text{seconds}
\]

So, the ball will take approximately **4.5 seconds** to fall back to the ground.

Would you like to explore further details or have any questions? Here are some related questions:

1. How high does the ball go?
2. What is the speed of the ball when it hits the ground?
3. How would the time change if the initial speed were doubled?
4. What would happen if the ball were launched at an angle instead of straight up?
5. How does air resistance affect the time it takes to reach the ground?

**Tip:** When dealing with projectile motion problems, always start by separating the motion into vertical and horizontal components if necessary.

Discover how long it takes for a ball launched straight up with an initial speed of 22 m/s to fall back to the ground. Follow a step-by-step solution using equations of motion and gravity acceleration. Suitable for Grades 9-12 students studying projectile motion in physics.

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