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A food packet is released from a helicopter which is trising steadily at ` 3 ms^(-1)`. After ` 3 seconds`, (i) what is the velocity of the packet? (ii) how far is it below the helicopter ? Take ` g= 9.8 m//s^2`. |
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Answer» Taking downwad motion of fodd parcet dropped from helicopter. ` Her, u =- 3 ms^(-1) , t = 3 s, a= + 9.8 m//s(-2)` (i) ` v=u + at =- 3 + (+ 9/8 ) 3 = + 26.4 ms^(-1)` This velocity is directed downwards. (ii) Downward distance coverd by packet in ` 3` seconds ` S= ut 1/2 at^2` ` =- 3 xx3 + 1/2 xx 9.8 xx3^2 = 35 .1 m` The heilcopter rises up in `3` seconds ` = 3 ms^(-1)` xx s= 9m` therfore, the distance of the food packet from the helecopter ` =35.1 + 9 44.1 m```. |
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