if u throw a ball up at 35m/s how long will its stay in the air

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.

Calculate the total time a ball will stay in the air when thrown vertically upward at 35 m/s using kinematic equations. The problem involves understanding equations of motion and projectile motion concepts. Suitable for students in Grades 9-10.

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