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a stone is thrown vertically upward with a velocity of 36m/s.after 2sec another ball is thrown vertically upwards, determine the initial velocity of 2nd ball so that they will meet at 30?

a stone is thrown vertically upward with a velocity of 36m/s.after 2sec another ball is thrown vertically upwards, determine the initial velocity of 2nd ball so that they will meet at 30?

Grade:12th pass

1 Answers

Vikas TU
14149 Points
6 years ago
Starting upward speed of first stone is u1=36m/s. After 2s it's speed v1=u1-gt=36–9.8X2=16.4. The stature accomplished by this stone=u1^2/2g=(36)^2/(2X9.8)=66.12m. Presently, as of now it begin openly and second stone begin to climb with it's underlying speed u2=9.8m/s. Beginning FROM THIS MOMENT assume they meet each other at time t. The separation shrouded by first stone in this time t is y1=(1/2)gt^2. The separation secured by second stone is y2=u2t-(1/2)gt^2=30u-4410. Presently, y1+y2=19.6=4.9t^2+9.8t-4.9t^2. 
 

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