a stone is projected vertically with a velocity of 20 meters/second ,two second later a second stone is similarly projected with the same velocity. when the two stones meet the second one is rising at a velocity of 10meter/second neglecting the air resistance calculate the:length of time the second stone is in motion before they met.
calculate the velocity of the first stone when they met (take g=10m/s
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Explanation:both the balls are thrown vertically upward with 20m/s at an interval of 2 sec
So time to reach maximum height= 2sec using 2nd equation of motion
So at the 3rd second of 1st ball both the balls will pass
The 1st ball will be moving with 10 m/s downward and 2nd ball with with 10 m/s upward
Velocith of 1st ball= 10 m/s downward
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