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 )
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Answered by
161
want to know - are u putting every question of the chapter here?
the ball A thrown upwards: let us say is thrown with speed u upwards.
height it reaches = h
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.
the ball A thrown upwards: let us say is thrown with speed u upwards.
height it reaches = h
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.
Answered by
20
The correct answer is (c) vA = vB Explanation: The ball 'A' thrown vertically upward will return back to the initial point with the same thrown speed but its direction will be downwards. This return speed is exactly the same as the speed of ball 'B'. So both of them will hit the ground with equal velocity.
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