Answer
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Hint Since the ball is thrown at 2 seconds interval, for both the balls to be in the air at the same time, the balls has to be in the air for more than 2 seconds. The ball when reaching the hand will have a final velocity equal to the initial velocity but opposite in direction.
Formula used: In this solution we will be using the following formulae;
\[v = u + at\] where \[v\] is the final velocity of an accelerating both (such as the one in free fall), \[u\] is the initial velocity, \[a\] is the initial velocity and \[t\] is the time elapsed.
Complete Step-by-Step solution:
The man throws the balls with the same velocity, hence the equation that applies to one ball applies to the others. Now, since he throws it at intervals of two seconds, the ball is to be in the air for more than two seconds.
The velocity the ball must be thrown at is the initial velocity, and the final velocity after two seconds is the velocity of the ball when it returns to the hand. In free fall, this final velocity is equal to the initial velocity but opposite in direction. Hence using the equation
\[v = u + at\] where \[v\] is the final velocity of an accelerating both (such as the one in free fall), \[u\] is the initial velocity, \[a\] is the initial velocity and \[t\] is the time elapsed, we can have
\[u = - u + gt\]
Hence, by inserting known values, we have
\[ \Rightarrow u = 9.8m/s\]
Hence, the initial velocity must be greater than \[9.8m/s\]
The correct option is B
Note: For clarity, the equation \[u = - u + gt\] has made the downward direction positive and the upward negative. This is a matter of choice, and hence it can be done the other way around as in
\[ - u = u - gt\]
Hence,
\[ - 2u = - gt\]
\[ \Rightarrow 2u = gt\]
Which is the same as above.
Formula used: In this solution we will be using the following formulae;
\[v = u + at\] where \[v\] is the final velocity of an accelerating both (such as the one in free fall), \[u\] is the initial velocity, \[a\] is the initial velocity and \[t\] is the time elapsed.
Complete Step-by-Step solution:
The man throws the balls with the same velocity, hence the equation that applies to one ball applies to the others. Now, since he throws it at intervals of two seconds, the ball is to be in the air for more than two seconds.
The velocity the ball must be thrown at is the initial velocity, and the final velocity after two seconds is the velocity of the ball when it returns to the hand. In free fall, this final velocity is equal to the initial velocity but opposite in direction. Hence using the equation
\[v = u + at\] where \[v\] is the final velocity of an accelerating both (such as the one in free fall), \[u\] is the initial velocity, \[a\] is the initial velocity and \[t\] is the time elapsed, we can have
\[u = - u + gt\]
Hence, by inserting known values, we have
\[ \Rightarrow u = 9.8m/s\]
Hence, the initial velocity must be greater than \[9.8m/s\]
The correct option is B
Note: For clarity, the equation \[u = - u + gt\] has made the downward direction positive and the upward negative. This is a matter of choice, and hence it can be done the other way around as in
\[ - u = u - gt\]
Hence,
\[ - 2u = - gt\]
\[ \Rightarrow 2u = gt\]
Which is the same as above.
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