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.

Respuesta :

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:

[tex]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[/tex]

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

[tex]h'=0\\-9.8*t+19.6=0\\t=19.6/9.8=2[/tex]

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

[tex]h(2)=-4.9*2^{2}+19.6*2-12.6\\\\h(2)=7[/tex]

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

The maximum height above the cliff top is 7 meters.

We have given that the equation,

[tex]h=-4.9*t^2+19.6*t-12.6[/tex]

What is the derivative of function?

derivative is the rate of change of a function with respect to a variable.

Here,rate of change of height and  variable is time and

Therefore the derivative of given function is,

[tex]h=-4.9*t^2+19.6*t-12.6[/tex]

[tex]h'=-9.8t+19.6[/tex]

We have to find t when the ball reaches the maximum height

[tex]h'=0\\-9.8t+19.6=0[/tex]

[tex]t=\frac{19.6}{9.8}=2[/tex]

we use the original equation to find the height after 2 seconds

[tex]h(2)=-4.9*2^2+19.6*2-12.6\\h(2)=7[/tex]

Therefor we get, the maximum height above the cliff top is 7 meters.

to learn more about the maximum height visit:

https://brainly.com/question/12446886