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From a balloon ascending with a velocity...

From a balloon ascending with a velocity of `9.8 ms^(-1)` , a stone was dropped and it reached the ground in 11s . How high was the balloon when the stone was dropped and with what velocity did it hit the ground ? (g `9.8 ms^(-2)` ) .

Text Solution

Verified by Experts

Given u `= -9.8 ms^(-1)`
t = 11 s , h = ? `V_(B) = ?`
By using the relation
h = `-ut + (1)/(2) "gt"^(2)` , we get
`h = -9.8 xx 11 + (1)/(2) xx 9.8 xx (11)^(2)`
`i.e. h = -107.8 + 592. 9`
i.e. h = 485.1 m
hence height of the balloon was 485.1 m above the ground .
`V_(B) = -u + "gt"`
`V_(B) = -9.8 + 9.8 xx 11`
`V_(B) = 98 ms^(-1)`
The stone hits the ground at `98 ms^(-1)`
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