A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0m/s when the hand is 1.50m 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.)

To find the time the ball is in the air before it hits the ground, we can use the kinematic equation:

h = ut + (1/2)gt^2

Where:
h = height or displacement
u = initial velocity
g = acceleration due to gravity
t = time

In this case, we know the following:
u = 19.0 m/s
h = 1.50 m
g = -9.8 m/s^2 (taking downward as the negative direction)

We need to find the time it takes for the ball to hit the ground, so h will be equal to 0 when the ball hits the ground.

0 = (19.0) t + (1/2)(-9.8) t^2

Simplifying the equation, we have:

0 = 19t - 4.9t^2

Rearranging the terms, we get the quadratic equation:

4.9t^2 - 19t = 0

Now we can solve for t by factoring out t:

t(4.9t - 19) = 0

This equation gives us two solutions: t = 0 (which is the initial time when the ball is thrown) and:

4.9t - 19 = 0

Solving for t:

4.9t = 19
t = 19 / 4.9 ≈ 3.88 seconds

So, the ball is in the air for approximately 3.88 seconds before it hits the ground.

To find the time it takes for the ball to hit the ground, we need to use the equation for vertical motion:

h = v0*t + (1/2) * g * t^2

Where:
- h is the height
- v0 is the initial velocity
- g is the acceleration due to gravity
- t is the time

We are given:
- v0 = 19.0 m/s (the initial velocity)
- h = 1.50 m (the initial height)

The acceleration due to gravity, g, is usually taken as -9.8 m/s^2 (assuming upward as positive and downward as negative).

Let's solve for t.

hf=hi+vi*t-1/2 g t^2

hf=0, hi=1.5 , vi=19

solve for t. It will be a quadratic.