Question

Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3 s with the same initial velocity of 35 m/s, then these balls collide at a height of _________ m.
(take \(g = 10 \text{ m/s}^2\))

Hint:

For relative motion under gravity, since both objects have the same acceleration (\(g\)), the relative acceleration is zero. The relative velocity is constant until they collide.

Answer 1

Correct Answer: 50

Step 1: Understanding the Concept:
This is a kinematics problem involving motion under gravity. The two balls collide when their vertical positions (heights) are equal at the same instant in time.
Step 2: Key Formula or Approach:
Displacement formula: \(h = ut - \frac{1}{2}gt^2\).
If the first ball is thrown at \(t = 0\), the second ball is thrown at \(t = 3 \text{ s}\).
Step 3: Detailed Explanation:
Let the collision occur at time \(t\) after the first ball is thrown.
Height of the first ball: \(h_1 = 35t - \frac{1}{2}(10)t^2 = 35t - 5t^2\).
Height of the second ball (travel time is \(t-3\)):
\[ h_2 = 35(t-3) - \frac{1}{2}(10)(t-3)^2 \]
At collision, \(h_1 = h_2\):
\[ 35t - 5t^2 = 35t - 105 - 5(t^2 - 6t + 9) \]
\[ -5t^2 = -105 - 5t^2 + 30t - 45 \]
\[ 30t = 150 \implies t = 5 \text{ s} \]
Now, find the collision height:
\[ h = 35(5) - 5(5)^2 \]
\[ h = 175 - 125 = 50 \text{ m} \]
Step 4: Final Answer:
The balls collide at a height of 50 m.