Question

A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by the function h(t)=-16t^2+20t+2. What is the ball's maximus height? How long does it take the ball to reach its maximum height? Round to the nearest hundredth, if necessary.

A. Reaches a max height of 8.25 feet after 1.25 seconds
B. Reaches a max height of 8.25 feet after 0.63 seconds
C. Reaches a max height of 0.16 feet after 1.34 seconds
D. Reaches a max height of 0.32 feet after 1.34 seconds.
I will give medal thanks!

Answer 1

What is the vertex of the parabola
\[

h(t)=-16t^2+20t+2.
\]

Answer 2

Or if you know derivative, what is the derivative of the above function?

Answer 3

Look at the graph

Answer 4

Without Calculus: We can write
\[h(t)=2-16 \left(t^2-\frac{5
t}{4}+\frac{25}{64}-\frac{2
5}{64}\right)=\\-16
\left(t^2-\frac{5
t}{4}+\frac{25}{64}\right)+
\frac{25}{4}+2=\\
-16
\left(t-\frac{5}{8}\right
)^2+\frac{33}{4}
\]

Answer 5

So, it attains it maximum height at t=5/8=.625 second and the maximum height is
33/4=8.25 ft

Answer 6

well that jerk @-_llama_- never gave the medal he promised..... this helped me out so thank you @eliassaab