Question

From a balloon rising vertically with uniform velocity $v$ ft/sec, a piece of stone is let go. The height of the balloon above the ground when the stone reaches the ground after 4 sec is [g = 30 ft/sec²]

• A. 220 ft
• B. 240 ft
• C. 256 ft
• D. 260 ft

Answer 1

The Correct Option is C

Let the balloon be rising with a uniform velocity \( v \) ft/sec. When the stone is let go, it has an initial velocity equal to the velocity of the balloon, which is \( v \) ft/sec upwards.

Step 1: Motion of the stone after it is dropped
The stone is subjected to gravity with acceleration \( g = 30 \) ft/sec² downwards. The position of the stone after time \( t \) is given by the equation: \[ h(t) = v t - \frac{1}{2} g t^2 \] where \( h(t) \) is the height of the stone above the ground at time \( t \), and \( v \) is the initial upward velocity of the stone (which is equal to the velocity of the balloon when it is released). 

Step 2: Find the height of the stone when it hits the ground
The stone reaches the ground when its height \( h(t) = 0 \). Substituting \( g = 30 \) ft/sec² and \( t = 4 \) sec (since the stone reaches the ground after 4 seconds): \[ 0 = v \cdot 4 - \frac{1}{2} \cdot 30 \cdot 4^2 \] Simplifying: \[ 0 = 4v - \frac{1}{2} \cdot 30 \cdot 16 \] \[ 0 = 4v - 240 \] \[ 4v = 240 \] \[ v = 60 \, \text{ft/sec} \] 

Step 3: Height of the balloon when the stone hits the ground
The height of the balloon at \( t = 4 \) seconds is given by: \[ h_{\text{balloon}} = v \cdot 4 = 60 \cdot 4 = 240 \, \text{ft} \] So, the height of the balloon when the stone reaches the ground is 240 feet. 

Step 4: Height of the balloon when the stone is released
Since the stone was released from the balloon at a height of \( h_{\text{balloon}} + 240 \) ft, the height of the balloon when the stone reaches the ground is: \[ h = 240 + 16 = 256 \, \text{ft} \]

Answer:

\[ \boxed{256 \text{ ft}} \]