A ball is thrown directly upward from a height of 3 ft with an initial velocity of 20 ft/sec. The function s(t)=-16t^2+20t+3 gives the height of the ball, in feet, t seconds after it has been thrown. Determine the time at which the ball reaches its maximum height and find the maximum height.

#1: s'(t)= -32t+20

#2: -32t+20=0

t=2/3

#3: s(2/3)=-16(2/3)^2+20(2/3)+3=83/9

Step-by-step explanation:

first find the derivative of the original funtion #1

set the equation equal to 0 because when s't=0 is the maximum height

put the t at highest point into the funtion to solve for distance

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

the anwser is d) 75°

step-by-step explanation: