A boy stands on the edge of a cliff of height 60 m. He throws a stone vertically upwards so that its distance, h m, above the cliff top is given by: h = 20t - 5t2   a) Calculate the maximum height of the stone above the cliff. b) Calculate the time which elapses before the stone hits the beach. (It just misses the boy and the cliff on the way down.) c) Calculate the speed with which the stone hits the beach.

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`h = 20t-5t^2`

Thus is related to maxima and minima in derivatives.

For maximum h then `(dh)/dt = 0`

`h = 20t-5t^2`

`(dh)/dt = 20-10t`

When `(dh)/dt = 0`

`20-10t = 0`

         t = 2

If h is a maximum at `t=2` then `(d^2h)/dt^2 <...

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`h = 20t-5t^2`

Thus is related to maxima and minima in derivatives.

For maximum h then `(dh)/dt = 0`

 

`h = 20t-5t^2`

`(dh)/dt = 20-10t`

 

When `(dh)/dt = 0`

 

`20-10t = 0`

         t = 2

 

If h is a maximum at `t=2` then `(d^2h)/dt^2 < 0` at t=2

`(dh)/dt = 20-10t`

(d^2h)/dt^2 = -10 <0

 

So h has a maximum at t=2

 

Maximum height above cliff `= 20*2-5*2^2 = 20m`

So the maximum height above cliff is 20m.

 

Since the editors can answer only one question i will left the other two for you.  But you can use the following guides.

  • When stone hits the beach what is h. h is (+) in upwards.you should remember that here.
  • The derivative of h function will give you the velocity function. Then its a matter of substituting the time.

 

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