To determine how long the ball will stay in the air, we can use the equations of motion under constant acceleration due to gravity. The key formula is:

$v = u + at$

Where:

• $v$ is the final velocity (which will be $0$ m/s at the highest point),
• $u$ is the initial velocity ($35$ m/s),
• $a$ is the acceleration due to gravity ($-9.8$ m/s², negative because it's acting downwards),
• $t$ is the time.

Step 1: Time to reach the highest point

At the highest point, the ball's velocity is 0, so:

$0 = 35 + (-9.8) t$

Solving for $t$:

$t = \frac{35}{9.8} \approx 3.57 \, \text{seconds}$

This is the time it takes to reach the highest point.

Step 2: Total time in the air

Since the time to go up is equal to the time to come down, the total time the ball stays in the air is:

$\text{Total time} = 2 \times 3.57 \approx 7.14 \, \text{seconds}$

So, the ball will stay in the air for approximately 7.14 seconds.

Would you like more details, or do you have any questions?

Here are 5 related questions to expand on this:

1. What is the maximum height the ball reaches?
2. How would the total time change if the initial velocity was 50 m/s?
3. What would happen if the ball was thrown from a height above the ground?
4. How do we calculate the speed of the ball after 2 seconds?
5. What if the gravity was on another planet, like the Moon?

Tip: When working with motion equations, always keep track of the signs of velocities and accelerations.