A ball is thrown upward and its height after t seconds can be described by formula f(t)=−10t2+24t+5.6. Find the maximum height the ball will reach.

Respuesta :

Answer:

The maximum height is 20.

Step-by-step explanation:

The height of the ball is given by [tex]f(t)= -10t^{2} +24t+5.6[/tex].

Differentiating the above equation, with respect to t, we get [tex]\frac{d f(t)}{dt} = -20t + 24[/tex].

At the maximum height the derivative of the function needs to be 0.

Hence, [tex]\frac{d f(t)}{dt} = 0[/tex] gives [tex]t = \frac{24}{20} = \frac{6}{5}[/tex].

The maximum height is [tex]f(\frac{6}{5} )= -10(\frac{6}{5} )^{2} +24(\frac{6}{5} )+5.6 = 20[/tex]