The height(h) of a stone, in meters, thrown into the air can be modeled by the equation h=-4.9t2+20t+10, where t represents time in seconds. How many seconds will it take for the stone to hit the ground (h=0) after it is thrown into the air? Round your answer to the tenths place.
Don't let the 'h's and 't's fool you or that negative coefficient on the t^2 term. It's still in a standard quadratic form: y = ax^2 + bx + c.
So, in h = -4.9·t^2 + 20·t + 10, a=-4.9, b=20, c=10
Of the different methods to factor a quadratic expression, this one seems to just call for the quadratic formula and a calculator or pencil & paper. With that b=20 you may think, complete the square but that -4.9 out front still makes this ugly.
Plug in the a, b, c values we identified and watch your signs: [-b ± √(b^2 - 4ac)] / 2a
As you plug and chug, you should get something close to these: t= 4.4/(-9.8), t=-44.4/(-9.8)
While we expect two solutions for h=0 in a quadratic, let's check back in with reality. In the problem, we can assume we're talking about t>0 seconds. So, we can ignore the first solution because t<0 or negative. Solving the other solution, you should get t≈4.5 sec.
Does this make sense? Well you can plug it back in the original formula to check. That's always best. You can also do some common sense looks at the problem using your quadratic expertise! Did you notice that if t=0, h=10? It's like someone's standing on a 10m high cliff to throw the stone up but we still want to know when it hits the ground 10m below. Quadratics (Parabolics) are symmetrical but the left part of this curve is cut-off i.e. if 't' were negative. So we'd expect this stone to hit it's peak height at something a bit less than half of 4.5sec. Drawing a picture really helps. Estimating 'h' at t=1, 2, 3 is easy enough (I rounded -4.9 to -5): h(t=1)≈25m, h(t=2)≈30m, h(t=3)≈25m √ The stone is hitting it's peak height somewhere around 2 sec which is less than half of 4.5sec. All looks reasonable.