Given that a ball is tossed from a 100 foot building upward with an initial velocity of 20 feet per second, the equation to model the situation is h(t)=−16t^2+20t+100 Answer the questions to follow, rounding to the nearest tenth. how high will the ball reach?

when will it hit the ground

Let y be the height above the ground for the ball. We can then write the equation as

y = -16t2 -22t + 130

If t = 0, then y = 130 feet as we are told.

We want to know when the ball strikes the ground, or when y = 0. So, set y = 0 and solve the quadratic for t.

-16t2 -22t + 130 = 0

We can divide the equation through by -2 to get

8t2 + 11t - 65 = 0

Unfortunately, there is no easy way to factor this equation. We go to the quadratic formula.

t = [-11 +/- [112 - 4(8)(-65)]1/2}/{(2)(8)}

We can simplify this as

t = [-11 +/- √2201]/16

√2201 > 11, so we only keep the positive root, since time must be positive.

So,

t = [-11 + √2201]/16 = 2.244 sec

c

Step-by-step explanation: