A ball is thrown from an initial height of 4 ft with an initial upward velocity of 40 ft/s. The ball's height h (in feet) after t seconds is given by the following. H=4+40t-16t squared. Find all values of t for which the ball's height is 26 feet

Step-by-step explanation:

you need o solve 26=4+40t-16t^2

the equation becomes:

the equation becomes:-16t^2+40t+4-26=0

the equation becomes:-16t^2+40t+-22=0 or8t^2-20t+11=0

8t^2-20t+11=0, a=8 , b=-20 c=11 the discriminant =b^2-4ac=(-20)^2-4x8x11=48

t1=(20-squarootof48)/16 =13/16 =0.81 seconds t2=27/16=1.68 second I rounded my answers

2/5

step-by-step explanation: