At `t=0` a 1.0 kg ball is thrown from a tall tower with `vec(v)=(18 m//s) hat(i)+(24 m//s) hat(j)`. What is `DeltaU` of the ball-Earth system between `t=0` and `t=6.0s` ( still free fall ) ?

Answer

To solve the problem of finding the change in potential energy (ΔU) of the ball-Earth system between \( t = 0 \) and \( t = 6.0 \) seconds, we can follow these steps:

### Step 1: Identify the initial conditions
At \( t = 0 \), the ball is thrown with a velocity vector:
\[ \vec{v} = 18 \hat{i} + 24 \hat{j} \, \text{m/s} \]
The mass of the ball \( m = 1.0 \, \text{kg} \).

### Step 2: Determine the vertical motion
Since we are interested in the change in potential energy, we only need to consider the vertical component of the motion. The initial vertical velocity \( v_{0y} = 24 \, \text{m/s} \).

### Step 3: Calculate the vertical displacement after 6 seconds
Using the kinematic equation for vertical displacement \( y \):
\[ y = v_{0y} t + \frac{1}{2} a t^2 \]
where \( a = -g = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity). Substituting the values:
\[ y = (24 \, \text{m/s}) \cdot (6 \, \text{s}) + \frac{1}{2} (-9.8 \, \text{m/s}^2) \cdot (6 \, \text{s})^2 \]
Calculating:
\[ y = 144 \, \text{m} - \frac{1}{2} \cdot 9.8 \cdot 36 \]
\[ y = 144 \, \text{m} - 176.4 \, \text{m} \]
\[ y = -32.4 \, \text{m} \]
This negative sign indicates that the ball has fallen 32.4 meters below the initial height.

### Step 4: Calculate the change in potential energy (ΔU)
The change in potential energy is given by:
\[ \Delta U = U_f - U_i = mgh_f - mgh_i \]
Assuming the potential energy at the initial height \( h_i \) is zero (taking it as the reference point), we have:
\[ \Delta U = mg(-32.4 \, \text{m}) \]
Substituting the values:
\[ \Delta U = (1.0 \, \text{kg})(9.8 \, \text{m/s}^2)(-32.4 \, \text{m}) \]
Calculating:
\[ \Delta U = -317.52 \, \text{J} \]

### Final Answer
The change in potential energy \( \Delta U \) of the ball-Earth system between \( t = 0 \) and \( t = 6.0 \, \text{s} \) is approximately:
\[ \Delta U \approx -317.52 \, \text{J} \]