Aball is launched upward at 20 meters per second (m/s) from a 60 meter tall platform. the equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 20t + 60, where s is in meters. how long before the object hits the ground after launch?

Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .