Approved Tutor Answer8 Years agoSuppose when first stone is dropped the upward velocity of the balloon is u. Let at this time height of the balloon be h. We take origin on the ground. The equation of motion of the first stone is y1=h+ut-(1/2)gt^2. After 2s the velocity of the balloon is u+0.2X2=u+0.4. This is initial velocity of second stone in upward direction . The height from ground is h+uX2+(1/2)X0.2X4=h+2u+0.4. This will be initial position of 2nd stone. Then, equation of motion for second stone is y2=h+2u+0.4+(u+0.4)(t-2)-(1/2)g(t-2)^2. Therefore y2-y1 can be found. We have taken time from the dropping of first stone. Therefore we ,while calculating y2-y1, have to take t=3.5s.The answer is 50m with g=10m/s^2. and 49m for g=9.8m/s^2.
Remember: acceleration of 0.2 m/s^2 does not act on stones from the moment of dropping of the stones.
i think your answer must be wrong.
Approved Tutor Answer8 Years ago
Three identical balls are placed on a frictionless
horizontal surface touching each other. They stick to
each other because of adhesive. Another ball of same
radius and mass m is placed over the void created by
the three balls. Find the forces applied by the balls
kept on the floor to each other if the system remains
in equilibrium.
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A car loaded with water having total mass 1000 kg moves on a straight horizontal Road starting from rest under the action of a force of 100 Newton. The water spills throw a small hole in the bottom at a rate of 0.1 KG per second the velocity of cart after 300 seconds is nearly.
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Last Activity: 10 Months ago
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Last Activity: 3 Years ago
