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Step-by-Step Solution
Step 1: Position of the first ball (Ball A) at t = 2 s
Ball A is launched vertically upwards with an initial velocity 50\,\text{m/s} at t = 0\,\text{s} .
To find its distance travelled from t = 0\,\text{s} to t = 2\,\text{s} , use the equation of motion:
s = ut - \tfrac{1}{2} g t^2
Substituting u = 50\,\text{m/s} , g = 10\,\text{m/s}^2 , and t = 2\,\text{s} :
s = 50 \times 2 - \tfrac{1}{2} \times 10 \times (2)^2
s = 100 - 20 = 80\,\text{m}
Therefore, at t = 2\,\text{s} , Ball A has reached 80 m above the launch point.
Step 2: Velocity of the first ball (Ball A) at t = 2 s
To find the velocity of Ball A at t = 2\,\text{s} , use:
v = u - g t
Substituting u = 50\,\text{m/s} , g = 10\,\text{m/s}^2 , and t = 2\,\text{s} :
v_A = 50 - 10 \times 2 = 30\,\text{m/s}
Thus, at t = 2\,\text{s} , Ball A is moving upward at 30\,\text{m/s} .
Step 3: Launch of the second ball (Ball B)
Ball B is projected vertically upward at t = 2\,\text{s} with the same initial velocity of 50\,\text{m/s} . At this moment, Ball A is already 80 m above the ground and moving upward at 30\,\text{m/s} .
Step 4: Relative motion between the two balls
• Velocity of Ball A at t = 2\,\text{s} : 30\,\text{m/s}
• Velocity of Ball B at t = 2\,\text{s} (launch instant for B): 50\,\text{m/s}
Immediately after t = 2\,\text{s} , the relative velocity of Ball B with respect to Ball A is:
v_{\text{rel}} = v_B - v_A = 50 - 30 = 20\,\text{m/s}
At t = 2\,\text{s} , the separation between the two balls is 80\,\text{m} . Both continue under the same gravitational acceleration ( 10\,\text{m/s}^2 downward), so their relative acceleration is:
a_{\text{rel}} = a_B - a_A = (-10) - (-10) = 0
Since the relative acceleration is zero, their relative speed remains 20\,\text{m/s} .
Step 5: Time for Ball B to catch up Ball A
Ball B must cover the 80 m gap at a relative speed of 20\,\text{m/s} . Hence, the time taken after t = 2\,\text{s} is:
t_{\text{meet}} = \dfrac{\text{Relative distance}}{\text{Relative speed}} = \dfrac{80}{20} = 4\,\text{s}
Step 6: Absolute time of meeting
Ball B was launched at t = 2\,\text{s} . Thus, the total time from t = 0 when the balls meet is:
t_{\text{total}} = 2 + 4 = 6\,\text{s}
Therefore, the two balls meet at t = 6\,\text{s} .
Final Answer
The second ball meets the first ball at t = 6\,\text{s} .
