Question

If a stone thrown vertically upwards from a bridge with an initial velocity of \(5 \operatorname{ms}^{-1}\), strikes the water below the bridge in a time of 3 s, then the height of the bridge above the water surface is (Acceleration due to gravity = \(10 \operatorname{ms}^{-2}\))

• A. \(10 \operatorname{m} \)
• B. \(26 \operatorname{m} \)
• C. \(30 \operatorname{m} \)
• D. \(18 \operatorname{m} \)

Hint:

When solving problems involving vertical motion under gravity: 1. Choose a consistent coordinate system: Define a positive direction (e.g., upwards or downwards) and stick to it for all vector quantities (displacement, velocity, acceleration). 2. Identify knowns and unknowns: List all given values and what needs to be found. 3. Select the correct kinematic equation: Choose from the standard equations of motion \((v = u + at, s = ut + \frac{1}{2}at^2, v^2 = u^2 + 2as)\) based on the variables involved. 4. Pay attention to signs: Acceleration due to gravity is always directed downwards. If upwards is positive, \(a\) is negative; if downwards is positive, \(a\) is positive.

Answer 1

The Correct Option is C

Step 1: Define the coordinate system and identify the knowns.
Let's choose the point where the stone is thrown from the bridge as the origin \((y=0)\).
Let the upward direction be positive, and the downward direction be negative.
Given values:
Initial velocity, \(u = +5 \operatorname{ms}^{-1}\) (positive because it's upwards).
Time taken to strike the water, \(t = 3 \operatorname{s}\).
Acceleration due to gravity, \(a = -10 \operatorname{ms}^{-2}\) (negative because it acts downwards).
We need to find the height of the bridge above the water surface, let's call it \(h\).
When the stone strikes the water below the bridge, its final position (displacement from the origin) will be \(-h\) (since it's below the starting point). So, \(s = -h\). 
Step 2: Apply the appropriate kinematic equation.
We use the second equation of motion, which relates displacement, initial velocity, acceleration, and time: \[ s = ut + \frac{1}{2}at^2 \] Step 3: Substitute the values and solve for \(h\).
Substitute the known values into the equation: \[ -h = (5)(3) + \frac{1}{2}(-10)(3)^2 \] \[ -h = 15 + \frac{1}{2}(-10)(9) \] \[ -h = 15 - 5(9) \] \[ -h = 15 - 45 \] \[ -h = -30 \] \[ h = 30 \operatorname{m} \] Step 4: State the conclusion.
The height of the bridge above the water surface is 30 meters. The final answer is $\boxed{30 \operatorname{m}}$.