Answer:
(a) The ball will hit the ground after 3 seconds
(b) The maximum height is 6.125
Step-by-step explanation:
Given
[tex]h(t) = -2t^2 +5t +3[/tex]
Solving (a): When the frisbee will hit the ground?
To do this, we set h(t) to 0
So, we have:
[tex]h(t) = -2t^2 +5t +3[/tex]
[tex]-2t^2 +5t +3=0[/tex]
Expand
[tex]-2t^2 +6t-t +3=0[/tex]
Factorize
[tex]-2t(t - 3) -1(t - 3) = 0[/tex]
Factor out t - 3
[tex](-2t -1)(t - 3) = 0[/tex]
Split:
[tex]-2t -1= 0\ or\ t - 3 = 0[/tex]
Solve for t in both equations
[tex]-2t =1\ or\ t = 3[/tex]
[tex]t =-\frac{1}{2}\ or\ t = 3[/tex]
Time can't be negative; So:
[tex]t = 3[/tex]
Solving (b): How height the frisbee will go?
First, we calculate time to reach the maximum height
[tex]t = -\frac{b}{2a}[/tex]
Where:
[tex]h(t) = at^2 + bt + c[/tex]
By comparison:
[tex]a = -2,\ b =5,\ c =3[/tex]
So:
[tex]t = -\frac{b}{2a}[/tex]
[tex]t = -\frac{5}{2*-2}[/tex]
[tex]t = \frac{5}{4}[/tex]
[tex]t = 1.25[/tex]
So, the maximum height is:
[tex]h_{max} = -2 * 1.25^2 + 5 * 1.25 + 3[/tex]
[tex]h_{max} = 6.125[/tex]