Ball `A` is thrown up vertically with speed `10 m/s`. At the same instant another ball `B` is released from rest at height `h`. At time `t`, the speed of `A` relative to `B` is

Options

A. `10`
B. `10-2 g t`
C. `sqrt(10^(2)-2gh)`
D. `10- g t`

Answer

To solve the problem of finding the speed of ball A relative to ball B at time t, we can follow these steps:

### Step 1: Understand the Motion of Ball A
Ball A is thrown upwards with an initial speed of \( u_A = 10 \, \text{m/s} \). The acceleration acting on it is due to gravity, which acts downwards, so we can denote it as \( a = -g \) where \( g \approx 9.8 \, \text{m/s}^2 \).

Using the equation of motion:
\[ v_A = u_A + a t \]
Substituting the known values:
\[ v_A = 10 - g t \]

### Step 2: Understand the Motion of Ball B
Ball B is released from rest at height \( h \). Its initial speed is \( u_B = 0 \). Since it is in free fall, it also experiences acceleration due to gravity, which is \( a = -g \).

Using the equation of motion:
\[ v_B = u_B + a t \]
Substituting the known values:
\[ v_B = 0 - g t = -g t \]

### Step 3: Calculate the Relative Speed of A with Respect to B
The relative speed of A with respect to B is given by:
\[ v_{AB} = v_A - v_B \]
Substituting the expressions for \( v_A \) and \( v_B \):
\[ v_{AB} = (10 - g t) - (-g t) \]
This simplifies to:
\[ v_{AB} = 10 - g t + g t = 10 \, \text{m/s} \]

### Final Answer
Thus, the speed of ball A relative to ball B at time \( t \) is:
\[ \boxed{10 \, \text{m/s}} \]