A boy standing on the top of a tower of height 54 ft. throws a packet with a speed of 20 ft/s directly aiming towards his friend standing on the ground at a distance of 72 ft from the foot of the tower. The packet falls short of the person on the ground by `x xx(16)/(3)` ft. The value of x is

Answer

To solve the problem step by step, we will follow the physics principles involved in projectile motion.

### Step 1: Understand the problem
A boy throws a packet from the top of a tower that is 54 ft high, with an initial speed of 20 ft/s, aiming at a friend who is 72 ft away from the base of the tower. We need to find out how much the packet falls short of the friend.

### Step 2: Identify the parameters
- Height of the tower (h) = 54 ft
- Horizontal distance to the friend (d) = 72 ft
- Initial speed of the packet (u) = 20 ft/s
- Acceleration due to gravity (g) = 32 ft/s² (approximately, since 1 ft = 0.305 m and g = 9.8 m/s²)

### Step 3: Calculate the time of flight (t)
The packet is thrown vertically downwards, so we will use the equation of motion for vertical displacement:
\[ s = ut + \frac{1}{2} a t^2 \]

Here,
- \( s = 54 \) ft (the height of the tower)
- \( u = 0 \) ft/s (initial vertical velocity)
- \( a = g = 32 \) ft/s² (acceleration due to gravity)

Substituting the values:
\[ 54 = 0 \cdot t + \frac{1}{2} \cdot 32 \cdot t^2 \]
\[ 54 = 16t^2 \]
\[ t^2 = \frac{54}{16} \]
\[ t^2 = 3.375 \]
\[ t = \sqrt{3.375} \approx 1.84 \text{ seconds} \]

### Step 4: Calculate the horizontal distance traveled by the packet
Now, we can find out how far the packet travels horizontally during this time:
\[ \text{Horizontal distance} = u \cdot t \]
\[ \text{Horizontal distance} = 20 \cdot 1.84 \approx 36.8 \text{ ft} \]

### Step 5: Calculate how far the packet falls short
The packet was aimed at a friend who is 72 ft away from the base of the tower. The distance the packet actually traveled horizontally is approximately 36.8 ft.

To find how much it fell short:
\[ \text{Shortfall} = d - \text{Horizontal distance} \]
\[ \text{Shortfall} = 72 - 36.8 = 35.2 \text{ ft} \]

### Step 6: Relate the shortfall to the given expression
According to the problem, the packet falls short by \( x \cdot \frac{16}{3} \) ft. Therefore, we can set up the equation:
\[ 35.2 = x \cdot \frac{16}{3} \]

### Step 7: Solve for x
To find x, we rearrange the equation:
\[ x = 35.2 \cdot \frac{3}{16} \]
\[ x = \frac{105.6}{16} \]
\[ x = 6.6 \]

### Final Answer
The value of \( x \) is approximately 6.6.

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