Asked by Kharl
An object is released to fall from height of 100m. At the same time another object is projected vertically upwards with velocity of 40m /s.
(i) Calculate the time taken before the objects meet
(ii) At what height do the objects meet
(i) Calculate the time taken before the objects meet
(ii) At what height do the objects meet
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Answered by
Bot
(i) The time taken before the objects meet can be calculated using the equation:
t = (2h)/(u + v)
where h is the height of the object, u is the initial velocity of the object falling, and v is the initial velocity of the object projected upwards.
In this case, h = 100m, u = 0m/s, and v = 40m/s.
Therefore, t = (2*100)/(0 + 40) = 5 seconds.
(ii) The objects will meet at a height of 100m, since the object falling from a height of 100m will take 5 seconds to reach the ground, and the object projected upwards will take 5 seconds to reach a height of 100m.
t = (2h)/(u + v)
where h is the height of the object, u is the initial velocity of the object falling, and v is the initial velocity of the object projected upwards.
In this case, h = 100m, u = 0m/s, and v = 40m/s.
Therefore, t = (2*100)/(0 + 40) = 5 seconds.
(ii) The objects will meet at a height of 100m, since the object falling from a height of 100m will take 5 seconds to reach the ground, and the object projected upwards will take 5 seconds to reach a height of 100m.
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