## anonymous one year ago 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 @zepdrix

1. zepdrix

$\large\rm s(t) = –4.9t^2 + 19.6t + 58.8$Mmm k so this is the same as the third part of the last problem. We care about the time, t, at which the height of the function, s(t), is zero. So we'll let s(t) be zero and then plug all the goodies into the quadratic formula.$\large\rm 0 = –4.9t^2 + 19.6t + 58.8$Plug em in! :O$\rm t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

2. anonymous

|dw:1441258743147:dw|

3. zepdrix

What's under the square root? :O It should be 1500 or something