SOLUTION: A ball is thrown vertically upwards from the balcony of an apartment building. The ball falls to the ground. The height of the ball , h metres, above the ground after t seconds i

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A ball is thrown vertically upwards from the balcony of an apartment building. The ball falls to the ground. The height of the ball , h metres, above the ground after t seconds i      Log On


   

Question 140056: A ball is thrown vertically upwards from the balcony of an apartment building. The ball falls to the ground. The height of the ball , h metres, above the ground after t seconds is given by h=-5t^2 +15t +45.
What is the height of the balcony? 45m?
What is the maximum height reached by the ball?
When will the ballreach a height of 55m?
When will the ball hit the ground?
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
h=-5t^2 +15t +45

Yes, the height of the building is the height at t=0, which is 45 ft.

Maximum height of the parabola y=ax%5E2+%2Bbx+%2Bc occurs at x=-b%2F%282a%29
Maximum height at =t=-15%2F%282%2A-5%29=15%2F10=3%2F2 seconds.
Maximum height = -5%2A%283%2F2%29%5E2%2B15%2A%283%2F2%29%2B45 feet
Maximum height = -5%289%2F4%29%2B45%2F2%2B45=-45%2F4%2B45%2F2%2B45
Maximum height = -45%2F4%2B90%2F4%2B45=45%2F4%2B45=11.25%2B45+=56.25 feet

The ball reaches 55 feet when
55=-5t%5E2+%2B15t+%2B45
0=-5t%5E2%2B15t-10
0=-5%28t%5E2-3t%2B2%29+
0=-5%28t-1%29%28t-2%29+
t=1 and t=2 seconds (at t=1 sec going up, at t=2 seconds coming down!)

The ball hits the ground when h=0.

h=-5t^2 +15t +45
0=-5t^2 +15t +45
0=-5(t^2-3t-9)
Solve this by quadratic formula, since it does not factor. The answer will contain a radical.

R^2