from a point A,80m above the ground , the particle is projected vertically upwards with a velocity of 29.4ms-1. 5 seconds later, another particle is dropped from a point B,34.3m vertically below A . when and where one overtakes the other [take g=9.8ms-2]

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Darshil Patel · Answer

Please apply this equation h+34.3=29.4+(-1/2)gt^2 So t=8 sec H=45.7m...... h=1.6m...... Okay understand

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from a point A,80m above the ground , the particle is projected vertically upwards with a velocity of 29.4ms-1. 5 seconds later, another particle is dropped from a point B,34.3m vertically below A . when and where one overtakes the other [take g=9.8ms-2]

from a point A,80m above the ground , the particle is projected vertically upwards with a velocity of 29.4ms-1. 5 seconds later, another particle is dropped from a point B,34.3m vertically below A . when and where one overtakes the other [take g=9.8ms-2]

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from a point A,80m above the ground , the particle is projected vertically upwards with a velocity of 29.4ms-1. 5 seconds later, another particle is dropped from a point B,34.3m vertically below A . when and where one overtakes the other [take g=9.8ms-2]

from a point A,80m above the ground , the particle is projected vertically upwards with a velocity of 29.4ms-1. 5 seconds later, another particle is dropped from a point B,34.3m vertically below A . when and where one overtakes the other [take g=9.8ms-2]

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