A ball is thrown into the air with an initial upward velocity of 45 ft/s. its height (h) in feet after t seconds is given by the function h = -16t^2 + 45t + 6. after about how many seconds will the ball hit the ground?

Question

tkhunny · Answer

Set h = 0 and solve.  You should get two answers. One should seem unreasonable.  Go!

anonymous · Answer

i don't really understand what am i suppose to do..

anonymous · Answer

@tkhunny

nnesha · Answer

like he said set h=0
replace h by 0
bec when ball hit the ground h should be 0

anonymous · Answer

Just look at the vertical distance (the height of the ball, h) Well if you look at this picture, notice that the ball is at (0,0) at the beginning, and it is also at 0 at the end. No? Link to pic. http://hs-mathematics.wikispaces.com/file/view/Ball.jpg/219992022/620x380/Ball.jpg

anonymous · Answer

so that is why tkhunny told you to put in 0 for h.  
If you do that, and solve for x, you will get the value of x where the ball end at height 0.

tkhunny · Answer

h = -16t^2 + 45t + 6

This thing says, "I was thrown from a spot 6 ft off the ground."  That's what the '6' on the end is all about.  Try it for t = 0.

-16(0)^2 + 45(0) + 6 = 0 + 0 + 6 = 6  Yeah!!

Okay, when will it hit the ground?

-16t^2 + 45t + 6 = 0

Use all your algebra power to solve this quadratic equation.