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 equationh = − 4.9t^2 + 19.6t − 12.6.Calculate, to the nearest metre, the maximum height the ball achieves above the cliff top.
Accepted Solution
A:
Answer:7 metersStep-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.