f(t) = -4.9t² + 24t + 8
a) The maximum height of the ball is given by the value of the function of the vertex of its parabola
For any quadractic function g(x) = ax² + bx + c, its vertex is given by:
[tex](x_v,y_v)=(-\frac{b}{2a},-\frac{b^2-4ac}{4a})_{}[/tex]
Then, the maximum heigt of the ball is given by:
[tex]\begin{gathered} h_{\max }=-\frac{24^2-4\cdot(-4.9)\cdot8_{}}{4\cdot(-4.9)} \\ h_{max}\approx37.4\text{ m} \end{gathered}[/tex]
b) When f(t) = 0, we have:
[tex]\begin{gathered} -4.9t^2+24t+8=0 \\ t_{}=\frac{-24\pm\sqrt[]{24^2-4\cdot(-4.9)\cdot8}}{2\cdot(-4.9)} \end{gathered}[/tex]
The only positive value is 5.2 (rounded)
Therefore, the ball hit the ground after 5.2 seconds
