Question

An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground?

Answer 1

put s = 0

find t

Answer 2

\[0=-4.9t^2+19.6t+58.8\]

Answer 3

awsumm, solve the quadratic equation now

Answer 4

hint: all can be divided by 4.9