To solve this problem, we can focus on the vertical motion of the tennis ball because the time it takes to reach the ground depends only on the vertical component of the motion.

Known values:

• Initial height ($h_0$) = 50 cm = 0.50 m
• Initial vertical speed ($v_{y0}$) = 0 m/s (the ball is hit horizontally)
• Acceleration due to gravity ($g$) = 9.8 m/s²
• Final height ($h$) = 0 m (ground level)

Equation:

We will use the kinematic equation for vertical displacement:

$h = h_0 + v_{y0} t + \frac{1}{2} g t^2$

Since the final height $h = 0$, and $v_{y0} = 0$, the equation simplifies to:

$0 = 0.50 + \frac{1}{2} \cdot 9.8 \cdot t^2$

Solving for $t$:

$\frac{1}{2} \cdot 9.8 \cdot t^2 = -0.50$ $t^2 = \frac{2 \cdot 0.50}{9.8}$ $t = \sqrt{\frac{1.0}{9.8}}$

Now, let's calculate the time.The time it takes for the ball to reach the ground is approximately $0.32$ seconds.

Would you like any further details or explanations? Here are some related questions you might consider:

1. What would the time be if the height was 1 meter instead?
2. How does the horizontal speed affect the total distance covered?
3. What would happen if the ball was hit at an angle instead of horizontally?
4. How much vertical distance does the ball travel in the first 0.1 seconds?
5. What is the velocity of the ball just before it hits the ground?

Tip: Horizontal and vertical motions in projectile problems are independent of each other. Only the vertical motion is influenced by gravity.