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Step-by-Step Detailed Solution
Step 1: Understand the Problem
We have two balls projected vertically upward with the same initial velocity of 50\,\text{m/s} , but the second ball is thrown 2 s after the first. We need to find the time (measured from t=0 when the first ball was thrown) at which they meet.
Step 2: Velocity of the First Ball at t = 2\,\text{s}
After the first ball has been moving for 2 s, its velocity decreases due to gravity ( g = 10\,\text{m/s}^2 ). The velocity of an object under uniform acceleration (like gravity) is
v = u - gt
(taking upward as positive and g as 10\,\text{m/s}^2 downward).
So, for the first ball at t = 2\,\text{s} :
v_1 = 50 - (10 \times 2) = 50 - 20 = 30\,\text{m/s}.
Step 3: Noting the Launch of the Second Ball
The second ball is launched at t = 2\,\text{s} with an initial velocity of 50\,\text{m/s} (the same as the first ball's original velocity). At the instant the second ball is launched, the first ball's velocity is 30\,\text{m/s} upward.
Step 4: Relative Motion Approach
To find when the two balls meet, we can consider the relative motion between the two balls right after t = 2\,\text{s} . At this instant:
The first ball continues upward with velocity 30\,\text{m/s} .
The second ball is just launched with velocity 50\,\text{m/s} .
The relative acceleration of one ball with respect to the other is
a_\text{rel} = g - g = 0,
because both are under the same gravitational acceleration downward.
Step 5: Find the Relative Displacement They Need to Cover
By the time t=2\,\text{s} , the first ball has traveled a certain vertical distance. We can determine how much higher the first ball will travel relative to where it was at t=2\,\text{s} , until the velocities become the key factor. Another way is to notice the difference in their velocities and use the relative motion idea directly:
- The velocity of the first ball relative to the second ball right when the second ball is launched at t=2\,\text{s} is
v_\text{rel} = (30 - 50) = -20\,\text{m/s}.
(This would indicate the first ball is moving slower; however, it is easier to see the needed distance for them to converge in another consistent manner.)
Let's instead consider how much extra height the first ball has above the launch point of the second ball at t=2\,\text{s} .
In the given solution, they've used:
S = \frac{u^2 - v^2}{2g} = \frac{50^2 - 30^2}{2 \times 10} = \frac{2500 - 900}{20} = \frac{1600}{20} = 80\,\text{m}.
This S = 80\,\text{m} is the difference in the displacement each ball would have when considering one ball's velocity dropping from 50\,\text{m/s} to 30\,\text{m/s} under gravity. It effectively captures how much "extra distance" exists between the two states of velocity.
Step 6: Calculate the Relative Velocity
When the second ball is launched, its velocity is 50\,\text{m/s} , and the first ball's velocity is 30\,\text{m/s} . So the difference in their velocities at that moment is:
v_\text{rel} = 50 - 30 = 20\,\text{m/s}.
Step 7: Time for the Second Ball to Meet the First Ball
Since the relative acceleration is zero (both experience the same g ), the time to close the relative distance S = 80\,\text{m} with relative speed v_\text{rel} = 20\,\text{m/s} is
\Delta t = \frac{S}{v_\text{rel}} = \frac{80}{20} = 4\,\text{s}.
Step 8: Total Time From t = 0
The second ball was launched at t = 2\,\text{s} , and it takes an additional 4 s from that moment for the balls to meet. Therefore, the meeting time from t=0 is:
t_\text{meet} = 2 + 4 = 6\,\text{s}.
Final Answer
The two balls meet at t = 6\,\text{s} .
