The equation y = -6t^2 + 51t describes the height (in feet) of a projectile launched from the surface of Mars at 51 feet per second. In how many seconds will the projectile first reach 100 feet in height?
This means y = 100
\(-6t^2+51 = 100\)
\(-6t^2+51 =100 = 0\)
Apply quadratic formula to get the positive solution of \(\frac{17}{4}+\frac{\sqrt{201}}{12}\)
This means y = 100
\(-6t^2+51 = 100\)
\(-6t^2+51 =100 = 0\)
Apply quadratic formula to get the positive solution of \(\frac{17}{4}+\frac{\sqrt{201}}{12}\)