Answer:
See below
Step-by-step explanation:
First Problem
The ball hits the ground when [tex]h(t)=0[/tex], therefore:
[tex]h(t)=-4.9t^2+v_0t+h_0[/tex]
[tex]0=-4.9t^2+32.7t[/tex]
[tex]0=t(-4.9t+32.7)[/tex]
[tex]t=0[/tex] and [tex]t=\frac{32.7}{4.9}\approx6.67[/tex]
Since the ball is in the air before it hits the ground, [tex]t=6.67[/tex] (seconds) is the more appropriate choice.
Second Problem
The maximum height of the ball is determined when [tex]t=-\frac{b}{2a}[/tex], therefore:
[tex]t=-\frac{b}{2a}[/tex]
[tex]t=-\frac{32.7}{2(-4.9)}[/tex]
[tex]t=-\frac{32.7}{-9.8}[/tex]
[tex]t\approx3.34[/tex]
This means that the height of the ball is at its maximum after 3.34 seconds:
[tex]h(t)=-4.9t^2+32.7t[/tex]
[tex]h(3.34)=-4.9(3.34)^2+32.7(3.34)[/tex]
[tex]h(3.34)\approx54.55[/tex]
Thus, the answer is 54.55 (meters).
Third Problem
Refer to the second problem
Fourth Problem
[tex]h(t)=-4.9t^2+32.7t[/tex]
[tex]h(4.3)=-4.9(4.3)^2+32.7(4.3)[/tex]
[tex]h(4.3)\approx50.01[/tex]
Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).
Fifth Problem
The ball will be 24 meters off the ground when [tex]h(t)=24[/tex], therefore:
[tex]h(t)=-4.9t^2+32.7t[/tex]
[tex]24=-4.9t^2+32.7t[/tex]
[tex]0=-4.9t^2+32.7t-24[/tex]
[tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\frac{-32.7\pm\sqrt{(32.7)^2-4(-4.9)(-24)}}{2(-4.9)}[/tex]
[tex]t_1\approx0.84[/tex]
[tex]t_2\approx5.83[/tex]
Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)
