To solve the problem, we need to analyze the motion of the two balls and equate their heights to find the time $t$ when both are at the same distance from the ground.

Step 1: Equations of Motion

For Ball 1 (dropped from height $h$):

• Initial velocity: $u_1 = 0$ (dropped from rest)
• Displacement equation under gravity: $y_1(t) = h - \frac{1}{2} g t^2$

For Ball 2 (launched upwards with initial velocity $u$):

• Initial velocity: $u_2 = u$
• Displacement equation under gravity: $y_2(t) = u t - \frac{1}{2} g t^2$

Step 2: Equating Heights

At the time $t$, the heights of the two balls are equal: $y_1(t) = y_2(t)$ Substitute the expressions for $y_1(t)$ and $y_2(t)$: $h - \frac{1}{2} g t^2 = u t - \frac{1}{2} g t^2$

Cancel out the $-\frac{1}{2} g t^2$ terms on both sides: $h = u t$

Solve for $t$: $t = \frac{h}{u}$

Step 3: Verify and Match the Answer

From the calculations, the time $t = \frac{h}{u}$. This corresponds to option C.

Let me know if you need additional clarification! 😊

Related Questions:

1. How does gravity affect the motion of both balls in this scenario?
2. What would the equation for $y_2(t)$ look like if air resistance were considered?
3. How does the initial height $h$ impact the time $t$?
4. If Ball 2 were launched at an angle instead of vertically, how would the equations change?
5. What happens if $h = 0$? Will the balls meet at any point?

Tip:

Always check the signs of the terms in motion equations to ensure they align with the direction of motion (positive for upwards, negative for downwards).