Aball is launched into the sky at 19.6 feet per second from a 58.8 meter tall building. the equation for the ball’s height, h, at time t seconds is h = -4.9t2 + 19.6t + 58.8 . when will the ball strike the ground?

A ball is launched into the sky at 19.6 feet per second from a 58.8 meter tall building. " 

Is that a typo, or a real mixture of units? 

The equation is only good if the launch speed is actually 19.6 m/s, so I'll assume that. 

It will strike the ground when the ball's height = 0 m !!! 

-4.9t² +19.6t + 58.8 = 0 

You can always use the "quadratic formula" if you want to, but ... 

This set of numbers is beloved by physics teachers everywhere. Everything divides by -4.9 

t² - 4t -12 = 0 

...and factorizes easily: 

(t + 2)(t - 6) 

So t = 6 or t = -2 Disregard the negative solution. 

The ball will strike the ground at t = 6 seconds

Distance=slopeslope is -2/4which equals-1/2the distance between these two points is -1/2