A baseball thrown straight up into the air has height s(t)6+48t-16t^2 feet after t seconds.Its velocity at time t is given by v(t)=48-32t. At what time(s) will the ball be above the ground?

Question

A baseball thrown straight up into the air has height s(t)6+48t-16t^2 feet after t seconds.Its velocity at time t is given by v(t)=48-32t. At what time(s) will  the ball be above the ground?

anonymous · Answer

So he threw the ball from 6 ft above ground level?

anonymous · Answer

Are we using that reference as ground level?

anonymous · Answer

Yes to both

campbell_st · Answer

find t = 0  for distance the object starts 6 feet above the ground.... when is the distance next equal to zero... use the general quadratic formula 

$$(-48\pm \sqrt{48^2 - 4 \times 3 \times (-16)})/(2\times(-16)$$

campbell_st · Answer

t= 3.12 seconds.... is when the object hits the ground.... so the ball is above the ground 

0<=t<= 3.12 secs...

campbell_st · Answer

the ball also is on the ground at  t = -0.12 secs.... but I'll assume time needs to be positive in this question.  The centre of the flight is v(t) = 0  of t = 1.5 secs