A ball 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) = -4.9t^2+20t+60, where s is in meters. How long before the object hits the ground after launch?

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

1

x = 1

step-by-step explanation:

0 ÷ x = 0

0 ÷ 1 = 0