To determine the time the object spends in the air, we can use the following kinematic equation, which relates the final velocity ($v$), initial velocity ($v_0$), acceleration ($a$), and time ($t$):

$v = v_0 + a \cdot t$

Given:

• Final velocity $v = 49.0 \, \text{m/s}$
• Initial velocity $v_0 = 0 \, \text{m/s}$ (since the object starts from rest)
• Acceleration due to gravity $a = 9.8 \, \text{m/s}^2$

Now solve for time $t$:

$t = \frac{v - v_0}{a}$

Substituting the values:

$t = \frac{49.0 \, \text{m/s} - 0 \, \text{m/s}}{9.8 \, \text{m/s}^2} = \frac{49.0}{9.8} = 5.0 \, \text{seconds}$

Thus, the object spends 5.0 seconds in the air.

Would you like more details on the steps or have any further questions?

Here are 5 related questions:

1. How high is the bridge if the object falls for 5.0 seconds?
2. What would happen if the acceleration due to gravity was different, say on the moon?
3. How does air resistance affect the time of fall in real-world scenarios?
4. What would be the velocity after 2.5 seconds of free fall?
5. How do we calculate the distance fallen if we know the time?

Tip: The formula $v = v_0 + at$ is one of the basic kinematic equations and is useful when solving for time, velocity, or acceleration in uniformly accelerated motion.