A man standing on the edge of a cliff throws a ball straight up with a speed of 8 Km/h and then throws a ball down with a speed of 8 Km/h from the same position. What is the ratio of speeds with which the ball hit the ground.
Answers
Explanation:
Speed of a particle (magnitude of velocity) is the same at same height. So, when the man throws the ball with speed u, it will return to him with the same speed u. So when man throws the stone upward with speed u and downward with speed u, the initial velocity (from the cliff) will be the same.
Answer:
Speed of a particle (magnitude of velocity) is the same at same height.
So, when the man throws the ball with speed u, it will return to him with the same speed u.
v=u+at
⇒0=u−gt (final velocity on reaching the top is 0,negative sign due to downward acceleration)
⇒u=gt ---------(i)
While coming down v=u+at
⇒−v=0−gt (negative sign due to downward direction)
⇒v=gt ------(ii)
From (i) and (ii),
∣v∣=∣u∣
So when man throws the stone upward with speed u and downward with speed u, the initial velocity (from the cliff) will be the same. Hence, this will result in equal final velocity.
Hence the ratio is 1:1
this is the correct answer
Explanation:
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