The velocity in the time up to 10 seconds is given by

`10 + int (3-0.3t)dt = 3t - 0.15t^2 + 10`

The distance travelled in this time is

`int_0^10 (3t-0.15t^2 +10)dt = (3/2t^2 -0.05t^3 + 10t)|_0^10 = `

`3/2(100) - 0.05(1000) +100` `= 150 - 50 + 100 =...

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The velocity in the time up to 10 seconds is given by

`10 + int (3-0.3t)dt = 3t - 0.15t^2 + 10`

The distance travelled in this time is

`int_0^10 (3t-0.15t^2 +10)dt = (3/2t^2 -0.05t^3 + 10t)|_0^10 = `

`3/2(100) - 0.05(1000) +100` `= 150 - 50 + 100 = 200` m

The raindrop at that point has another 300m to fall

The velocity after 10s = 30 - 0.15(100) + 10 = 25m/s

It remains at this speed when t>10 as it has stopped accelerating

To fall 300m at this speed it takes 300/25 = 12s

**Therefore the raindrop takes 10 + 12 = 22s to fall to the ground**