Question

If two stones are thrown vertically upwards with their velocities in the ratio 2:5, then the ratio of the maximum heights attained by the stones is

• A. 4:25
• B. 1:5/2
• C. 6:15
• D. 8:20

Hint:

For projectile motion problems, remember the key relationships:
Maximum height \(H \propto u^2\) (for a given angle)
Time of flight \(T \propto u\)
Range \(R \propto u^2\)
These proportionality relations allow you to solve ratio problems very quickly without calculating the actual values.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When an object is thrown vertically upwards, its initial kinetic energy is converted into gravitational potential energy. At the maximum height, the velocity is momentarily zero. We can use the kinematic equation relating initial velocity, final velocity, acceleration, and displacement.
Step 2: Key Formula or Approach:
The relevant kinematic equation is: \[ v^2 = u^2 + 2as \] At the maximum height (\(H\)), the final velocity \(v = 0\). The acceleration is \(a = -g\). The displacement is \(s = H\). \[ 0 = u^2 + 2(-g)H \] Solving for the maximum height \(H\): \[ u^2 = 2gH \implies H = \frac{u^2}{2g} \] This shows that the maximum height is proportional to the square of the initial velocity (\(H \propto u^2\)).
Step 3: Detailed Explanation:
Let the initial velocities of the two stones be \(u_1\) and \(u_2\). We are given the ratio of their velocities: \[ \frac{u_1}{u_2} = \frac{2}{5} \] Let the maximum heights they reach be \(H_1\) and \(H_2\). Using the formula \(H = \frac{u^2}{2g}\), we can find the ratio of their heights: \[ \frac{H_1}{H_2} = \frac{u_1^2 / (2g)}{u_2^2 / (2g)} = \frac{u_1^2}{u_2^2} = \left(\frac{u_1}{u_2}\right)^2 \] Substitute the given ratio of velocities: \[ \frac{H_1}{H_2} = \left(\frac{2}{5}\right)^2 = \frac{4}{25} \] Step 4: Final Answer:
The ratio of the maximum heights attained by the stones is 4:25.