A ball is thrown straight up. Its height, h(in metres), after t seconds is given by h=-5t^2+10t+2. To the nearest tenth of a second, when is the ball 6m above the ground?
Explain why there are two answers.
you have to equate h to 6:
h = 6 ---->
5t^2+10t+2 = 6 --->
5 t^2 + 10 t - 4 = 0 --->
t = -1 +/- 3*sqrt[1/5]
Wow i have know idea But you said t Seconds. whats that?
t is a variable
It is at that height going up and coming down in its path throughout its motion.
The path is illustrated like an upside down u.
The times are after leaving what propelled it upward and as it falls back to earth via gravity.