A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 15m/s{\rm m}/{\rm s} when the hand is 2.0m{\rm m} above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

3.2s

To find the time it takes for the ball to hit the ground, we can use the kinematic equation that relates the time of flight to the initial velocity and the acceleration due to gravity.

The equation we can use is:

d = v₀t + 0.5gt²

Where:
- d is the final displacement (in this case, the height or distance above the ground)
- v₀ is the initial velocity
- t is the time of flight
- g is the acceleration due to gravity (approximated as 9.8 m/s²)

We are given:
- v₀ = 15 m/s (initial velocity)
- d = 2.0 m (the distance the hand is above the ground)
- g = 9.8 m/s² (acceleration due to gravity)

We need to solve for t.

Substituting the given values into the equation:

2.0 = (15)t + 0.5(9.8)t²

Now, we have a quadratic equation. Rearranging the terms:

0.5(9.8)t² + (15)t - 2.0 = 0

To solve this equation, you can use the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 0.5, b = 15, and c = -2.0.

Calculating the values:

t = (-15 ± √(15² - 4 * 0.5 * -2.0)) / (2 * 0.5)

Simplifying:

t = (-15 ± √(225 + 4)) / 1

t = (-15 ± √229) / 1

Now, you can calculate the two possible values for t by evaluating the square root:

t₁ = (-15 + √229) / 1

t₂ = (-15 - √229) / 1

Evaluate these expressions to find the two possible values for t.

By solving for t, you can determine how long the ball will be in the air before it hits the ground.

2.91s

idk man

2.99