An object drops from a cliff that is 150m high. The distance, d, in metres that the object has dropped at `t` seconds is modelled by `d(t)=4.9(t)^(2)`
- Find the rate at which the object hits the ground to the nearest tenth (answer should be in metres per second - m/s).
1 Answer | Add Yours
If the distance that the object has dropped at t seconds is modeled by
`d(t) = 4.9t^2` , the rate is modeled by the derivative of this function:
`r(t) = (dd(t))/dt = 9.8t` . The units of the rate are m/s.
The time at which the object hits the ground can be calculating by letting d = 150 m:
`150 = 4.9t`
`t = 150/4.9 = 30.61s`
At this time, the rate will be `r = 9.8*30.61 = 300` m/s.
The object hits the ground at the rate of 300 m/s.
We’ve answered 328,011 questions. We can answer yours, too.Ask a question