At the start of the basketball game, the referee tosses a ball for a jump ball. The equation that models the height of the ball is
h(t) = -16t2 + 24t + 5
During what time interval is the height of the ball above 9 feet? (Round your answer to the nearest tenth of a second.)
a) .6 seconds to 1.8 seconds
b) .4 seconds to 1.9 seconds
c) .2 seconds to 1.3 seconds
d) .5 seconds to 2.1 seconds

Question

At the start of the basketball game, the referee tosses a ball for a jump ball. The equation that models the height of the ball is 
h(t) = -16t2 + 24t + 5
During what time interval is the height of the ball above 9 feet? (Round your answer to the nearest tenth of a second.)
	a)	.6 seconds to 1.8 seconds
	b)	.4 seconds to 1.9 seconds
	c)	.2 seconds to 1.3 seconds
	d)	.5 seconds to 2.1 seconds

anonymous · Answer

Don't you think we have to take derivative of the equation first.???

eyust707 · Answer

no that would give you velocity

shane_b · Answer

It's already a position equation...

eyust707 · Answer

with the quadratic you will obtain two solutions

eyust707 · Answer

one time on the way up on on the way down

eyust707 · Answer

the difference between the two is the solution

shane_b · Answer

http://www.wolframalpha.com/input/?i=solve+for+t%3A+9%3D-16t^2+%2B+24t+%2B+5

eyust707 · Answer

nice

shane_b · Answer

I'm lazy

anonymous · Answer

the answer is c)

eyust707 · Answer

I guess the ref is 5 ft tall

shane_b · Answer

must be the WNBA

eyust707 · Answer

haha

this · Answer

there fore so the answer is C hun?

eyust707 · Answer

yep

shane_b · Answer

Definitely C.

campbell_st · Answer

you have a question where you need to solve fo t then h(t) = 9

so you have 

9 = -16t^2 + 24t + 5 

or 

-16t^2 + 24t - 4 = 0

so solve for t

lots of ways... but here is the long version using the general Quadratic formula

$$t = \frac{-24 \pm \sqrt{24^2 - 4\times(-16\times(-4)}}{2\times(-16)}$$

simplifying gives

$$t = \frac{-24 \pm \sqrt{576 - 256}}{-32}$$

after some further simplification you end up with

$$t = \frac{-3 \pm \sqrt{5}}{-4}$$

so you need to evaluate

$$t = \frac{-3 - \sqrt{5}}{-4} .... and ... t=\frac{-3 + \sqrt{5}}{-4}$$

its between these 2 time values that the ball is above 9 ft

campbell_st · Answer

t = 0.19  or rounding to 2 d.p    t = 0.2 sec and t = 1.3 sec

this · Answer

thnks guys a lot ya all r the best :)

callisto · Answer

Same question... I got the same numerical answer as @campbell_st, but I don't know why I was wrong.. The arrangement of the choices was different. http://openstudy.com/users/callisto#/updates/4fdca110e4b0f2662fd27820

shane_b · Answer

I don't see why it's wrong either...it's just a simple quadratic equation and you can plainly see the answer on the graph.

callisto · Answer

Very true indeed!!