SOLUTION: Solving a science application. Suppose that a ball is thrown upward with an initial velocity of 80 ft/s. (the initial velocity is the speed with which the ball leaves the throwers

Algebra ->  Pythagorean-theorem -> SOLUTION: Solving a science application. Suppose that a ball is thrown upward with an initial velocity of 80 ft/s. (the initial velocity is the speed with which the ball leaves the throwers       Log On


   

Question 119148: Solving a science application. Suppose that a ball is thrown upward with an initial velocity of 80 ft/s. (the initial velocity is the speed with which the ball leaves the throwers hand. The spped will then decrease as the ball rises.) If the ball is released at a height of 77ft, the height equation may be written as follows: h= -16t^2+80t+5, round to the nearest hundredth of a second
77= -16t^2+80t+5
this is all i got, now I am lost, can someone please help me, thank you
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Do you want to solve for t?



77=-16t%5E2%2B80t%2B5 Start with the given equation


0=-16t%5E2%2B80t%2B5-77 Subtract 77 from both sides.


0=-16t%5E2%2B80t-72 Combine like terms


Let's use the quadratic formula to solve for t:


Starting with the general quadratic

at%5E2%2Bbt%2Bc=0

the general solution using the quadratic equation is:

t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve -16%2At%5E2%2B80%2At-72=0 ( notice a=-16, b=80, and c=-72)




t+=+%28-80+%2B-+sqrt%28+%2880%29%5E2-4%2A-16%2A-72+%29%29%2F%282%2A-16%29 Plug in a=-16, b=80, and c=-72



t+=+%28-80+%2B-+sqrt%28+6400-4%2A-16%2A-72+%29%29%2F%282%2A-16%29 Square 80 to get 6400



t+=+%28-80+%2B-+sqrt%28+6400%2B-4608+%29%29%2F%282%2A-16%29 Multiply -4%2A-72%2A-16 to get -4608



t+=+%28-80+%2B-+sqrt%28+1792+%29%29%2F%282%2A-16%29 Combine like terms in the radicand (everything under the square root)



t+=+%28-80+%2B-+16%2Asqrt%287%29%29%2F%282%2A-16%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



t+=+%28-80+%2B-+16%2Asqrt%287%29%29%2F-32 Multiply 2 and -16 to get -32

So now the expression breaks down into two parts

t+=+%28-80+%2B+16%2Asqrt%287%29%29%2F-32 or t+=+%28-80+-+16%2Asqrt%287%29%29%2F-32


Now break up the fraction


t=-80%2F-32%2B16%2Asqrt%287%29%2F-32 or t=-80%2F-32-16%2Asqrt%287%29%2F-32


Simplify


t=5+%2F+2-sqrt%287%29%2F2 or t=5+%2F+2%2Bsqrt%287%29%2F2


So these expressions approximate to

t=1.1771243444677 or t=3.8228756555323


So our solutions are:
t=1.1771243444677 or t=3.8228756555323




So when the time is t=1.1771243444677 or t=3.8228756555323 the ball is at a height of 77 ft.