Question

A ball is thrown vertically upward. After t seconds, its height h, in feet, is given by the function LaTeX: h\left(t\right)=-16t^2+64th ( t ) = − 16 t 2 + 64 t. How many seconds will it take the ball to reach its maximum height?

Answer 1

Answer:

The time it will take for the ball to reach maximum height of 64 feet is 2 seconds

Step-by-step explanation:

The given equation for the height, h, in feet the ball reaches is given by the function;

h(t) = -16·t² + 64·t

To reach maximum height, $h_{max}$, we have;

At $h_{max}$, we have;

$\dfrac{\mathrm{d} h}{\mathrm{d} t} = 0$

$\dfrac{\mathrm{d} \left (-16\cdot t^{2} + 64\cdot t \right )}{\mathrm{d} t} = 0$

Which gives;

64 -32·t = 0

64 = 32·t

t = 64/32 = 2 seconds

Therefore, it will take 2 seconds for the ball to reach maximum height

The maximum height reached is given as follows;

h(2) = -16×2² + 64×2 = 64 feet

The time it will take for the ball to reach maximum height of 64 feet is 2 seconds.