A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with a speed V1 and the ball B hits the ground with a speed V2. We have,V1 < V2 or V1 = V2( answer should come V1 = V2 ) please answer in detail
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The ball A thrown upwards: let us say is thrown with speed u m/s upwards.
height it reaches = h m
then v² = u² + 2 a s
=> 0 = u² - 2 g h => u = √(2gh)
When the ball reaches the same point h, on the way down it will have a speed :
v² = 0 + 2 g h => v = √(2gh)
So the speeds of A and B will be same when they reach the ground, as they travel the same distance.
height it reaches = h m
then v² = u² + 2 a s
=> 0 = u² - 2 g h => u = √(2gh)
When the ball reaches the same point h, on the way down it will have a speed :
v² = 0 + 2 g h => v = √(2gh)
So the speeds of A and B will be same when they reach the ground, as they travel the same distance.
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