A person standing close to the edge on the top of a 160-foot building throws a baseball vertically upward. the quadratic function
s(t) = -16t^2+64t+160
models the ball's height above the ground, s(t), in feet, t seconds after it was thrown
a) after how many seconds does the ball reach its maximum height? What is the maximum height?
b) How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second
so I have found a) as 2 seconds and 224 feet but how do i figure out part b?

Question

A person standing close to the edge on the top of a 160-foot building throws a baseball vertically upward.  the quadratic function 
s(t) = -16t^2+64t+160
models the ball's height above the ground, s(t), in feet, t seconds after it was thrown 
a) after how many seconds does the ball reach its maximum height? What is the maximum height?
b) How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second
so I have found a) as 2 seconds and 224 feet but how do i figure out part b?

anonymous · Answer

the answer for part be is suppose to be 5.7 seconds

anonymous · Answer

Put s(t) =0 and then solve the quadratic.

anonymous · Answer

$s(t) = 0 =-16t^2+64t+160$

anonymous · Answer

From there we can divide thorugh to simplify and then factorise ( or quadratic formula) Does that make sense?

anonymous · Answer

i think so!

campbell_st · Answer

to find the max height... there are a couple of ways... differentiating... 

or 

find the line of symmetry, the max height is on the line of symmetry

for your question $$t = \frac{-b}{2a}$$

you have a = -16 and b = 64

so the max height occurs when t = 

$$t = \frac{-64}{2 \times -16} = 2$$

substitute  t=2 into the equation to find the max height. 

to find the time of flight.. let the height s(t) = 0 and solve

$$0 = -16(t^2 -4t - 10)$$

you need the general quadratic formula to find the time taken for it to hit the ground.