A juggler tosses a ball h = 2 m into the air with one hand and catches it with the other hand at the same level. Neglecting air resistance, how long is the ball in the air?

h=1/2gt

-2m=1/2gt
-4m=gt
-4m=(-9.8m/s^2)t
-4/(-9.8m/s^2)=t

To find the time the ball is in the air, we can use the equation of motion. The equation for the vertical position of an object in free fall is given by:

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

Where:
- h is the height of the ball (2 m)
- u is the initial velocity of the ball (when it is released, it goes upwards, so u is positive)
- g is the acceleration due to gravity (-9.8 m/s^2, taking downwards as negative)
- t is the time the ball is in the air (unknown)

Since the ball is tossed upwards, the initial velocity (u) is positive. At the highest point of the trajectory, the ball will have a velocity of zero before starting to fall back down.

Knowing this, we can rearrange the equation to solve for time (t):

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

Rearranging:

(1/2)*g*t^2 + u*t - h = 0

Since the ball starts at height h = 2m, we can substitute the values:

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

Now, the initial velocity is the same as the final velocity when the ball reaches the same level while coming back down. So the final velocity is also zero.

Substituting this into the equation, we get:

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

-4.9*t^2 - 2 = 0

Now, we can solve this quadratic equation for t using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a)

Where a = -4.9, b = 0, and c = -2.

Plugging in the values, we get:

t = (-0 ± sqrt(0^2 - 4*(-4.9)*(-2))) / (2*(-4.9))

Simplifying further:

t = (± sqrt(0 - 39.2)) / (-9.8)

Since we're measuring time, it can't be negative. So we take the positive square root:

t = sqrt(39.2) / 9.8

Calculating this:

t ≈ 2 s

Therefore, the ball is in the air for approximately 2 seconds.