A light shines from the top of a pole 50 ft. high A ball is dropped from the same height from a point 30 ft. away from the light . If the shadow of the ball moving at the rate of `100lambda ft//sec` along the ground `1//2`sec. later [Assume the ball falls a distance s= `16t ^(2) ft. in 't' sec ]` then `| lambda|` is :

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a pole of height 50 ft and a ball dropped from the same height, 30 ft away from the base of the pole. The ball falls according to the equation \( s = 16t^2 \) ft, where \( s \) is the distance fallen after \( t \) seconds. We need to find the value of \( |\lambda| \) given that the shadow of the ball moves at a rate of \( 100\lambda \) ft/sec after \( \frac{1}{2} \) seconds.

### Step 2: Determine the Distance Fallen by the Ball
Using the formula for distance fallen:
\[ s = 16t^2 \]
At \( t = \frac{1}{2} \) seconds:
\[ s = 16 \left( \frac{1}{2} \right)^2 = 16 \times \frac{1}{4} = 4 \text{ ft} \]

### Step 3: Find the Height of the Ball at \( t = \frac{1}{2} \)
The height of the ball after \( \frac{1}{2} \) seconds is:
\[ \text{Height} = 50 - s = 50 - 4 = 46 \text{ ft} \]

### Step 4: Set Up Similar Triangles
We can set up two similar triangles:
1. Triangle formed by the pole and the shadow of the ball.
2. Triangle formed by the distance from the light to the ball and the height of the ball.

Let \( x \) be the length of the shadow of the ball on the ground. By the properties of similar triangles, we have:
\[ \frac{50}{x + 30} = \frac{46}{x} \]

### Step 5: Cross-Multiply and Solve for \( x \)
Cross-multiplying gives:
\[ 50x = 46(x + 30) \]
Expanding and simplifying:
\[ 50x = 46x + 1380 \]
\[ 4x = 1380 \]
\[ x = \frac{1380}{4} = 345 \text{ ft} \]

### Step 6: Find the Rate of Change of the Shadow Length
To find the rate of change of the shadow length \( x \), we differentiate the relationship we established:
\[ \frac{50}{x + 30} = \frac{46}{x} \]
Differentiating both sides with respect to \( t \):
\[ \frac{d}{dt} \left( \frac{50}{x + 30} \right) = \frac{d}{dt} \left( \frac{46}{x} \right) \]

Using the quotient rule:
\[ -\frac{50 \cdot \frac{dx}{dt}}{(x + 30)^2} = -\frac{46 \cdot \frac{dx}{dt}}{x^2} \]

### Step 7: Substitute Known Values
At \( t = \frac{1}{2} \):
\[ \frac{dx}{dt} = \text{rate of shadow length change} \]
Substituting \( x = 345 \) ft into the equation:
\[ -\frac{50 \cdot \frac{dx}{dt}}{(345 + 30)^2} = -\frac{46 \cdot \frac{dx}{dt}}{(345)^2} \]

### Step 8: Solve for \( \frac{dx}{dt} \)
Canceling \( \frac{dx}{dt} \) from both sides (assuming it's not zero):
\[ \frac{50}{(375)^2} = \frac{46}{(345)^2} \]

Cross-multiplying gives:
\[ 50 \cdot (345)^2 = 46 \cdot (375)^2 \]

### Step 9: Find the Value of \( \lambda \)
We know that the speed of the shadow is given by:
\[ \frac{dx}{dt} = 100\lambda \]
Setting the two expressions for speed equal gives:
\[ 100\lambda = \text{calculated speed} \]

### Step 10: Calculate \( |\lambda| \)
From the calculations, we can find the value of \( \lambda \) and take the absolute value.