a ball thrown upward reaches a point P on the path or the end of 4 seconds and highest q at the end of 12 sec .after how many sec from the start will it reach the point p again
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20 seconds ..
First it goes above then at that reference of motion it takes 12 seconds as it is mentioned
But the total height of the motion of that particle is 720 meters ...and point P is located at 400 meters from the lowest point possible ..so 720-400 = 320 .
Now . 320 = 5t^2
64 = t^2
Which is 8 ..so the total time comes to be 12 + 8 seconds ..so as. 20 seconds
First it goes above then at that reference of motion it takes 12 seconds as it is mentioned
But the total height of the motion of that particle is 720 meters ...and point P is located at 400 meters from the lowest point possible ..so 720-400 = 320 .
Now . 320 = 5t^2
64 = t^2
Which is 8 ..so the total time comes to be 12 + 8 seconds ..so as. 20 seconds
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