A baseball is thrown straight up, reaches a height of 4.9 m, and is caught at the point it was thrown from. The ball was in the air for a total of how many seconds?

a) 96 s
b) 0.71 s
c) 1.0 s
d) 0.50 s
e) 2.0 s

Yeah A is not right.

To find the answer, we need to use the kinematic equation for vertical motion:

\(h = v_0t + \frac{1}{2}gt^2\)

Where:
h = height (4.9 m)
\(v_0\) = initial velocity (when the ball is thrown straight up, the initial velocity is positive)
t = time in the air
g = acceleration due to gravity (-9.8 m/s^2)

Since the ball is caught at the point it was thrown from, the final height is the same as the initial height. Therefore, we can set h = 0 and solve for t:

\(0 = v_0t + \frac{1}{2}gt^2\)

Since the ball was thrown straight up, the initial velocity is positive. We know that \(v_0\) = 0 m/s because the ball reaches its highest point momentarily and then starts to come down.

Thus, the equation becomes:

\(0 = \frac{1}{2}gt^2\)

Taking away the common factor of \(t^2\), we have:

\(0 = \frac{1}{2}g \times t \times t\)

\(0 = \frac{1}{2} \times (-9.8) \times t \times t\)

Simplifying the equation, we have:

\(0 = -4.9 \times t \times t\)

Now, divide both sides of the equation by -4.9:

\(0 = t \times t\)

\(0 = t^2\)

From this, we can conclude that \(t = 0\).

Therefore, the baseball was in the air for 0 seconds.

The correct answer is b) 0.71 s.

Note: The kinematic equation assumes idealized conditions and neglects factors such as air resistance.