Question

The elevation of a rock thrown from a bridge can be modeled by the expression -16t^2+80t+96 where t is the number of seconds since the rock was thrown and the elevation is the number of feet above water. What is the maximum height the rock will reach and how many seconds will it take for the rock to hit water again?

Answer 1

Maximum height is found by differentiating the elevation expression and equating to zero to find the value of t that leads to maximum height

First differentiation ;$d/dt( - 16t^2+80t+96) =-32t+80$

Equating the expression to zero :$-32t+80=0$

$32t=80$

$t=2.5$

To check if it's a maximum we find the second derivative which should be less than zero for a maximum point ;

Second derivative ;$d^2h/dt^2=d/dt( - 32t+80)=-32$ which is less than zero.

Maximum height reached is therefore at t=2.5 which is;$h=-16(2.5^2)+80(2.5)+96=196ft$

Time taken to hit water =2.5 seconds