Question

A ball is launched into the air from below a cliff, such that after t seconds its height above the cliff top is hmetres, and is given by the equation
h =  − 4.9t^2 + 19.6t − 12.6.

Calculate, to the nearest metre, the maximum height the ball achieves above the cliff top.

Answer 1

Answer:

7 meters

Step-by-step explanation:

In order to find the answer we need to calculate the first derivative of the function as follows:

$h=-4.9*t^{2}+19.6*t-12.6\\h'=-4.9*2*t^{2-1}+19.6*1-0\\h'=-9.8*t+19.6$

Now, for obtaining the time 't' when the ball reaches the maximum height:

$h'=0\\-9.8*t+19.6=0\\t=19.6/9.8=2$

Finally, we use the original equation for determining the height after 2 seconds:

$h(2)=-4.9*2^{2}+19.6*2-12.6\\\\h(2)=7$

In conclusion, the maximum height above the cliff top is 7 meters.