If a ball is thrown straight up with an initial velocity of 29.4 m/s, the equation for the height "h" is given by h = -4.9t2 + 29.4t. When does the
ball return to the ground?

6 sec

Step-by-step explanation:

when ball return to the ground, h is 0

0=at^2 +bt

0 = -4.9t^2+29.4t

factor: -4.9 times -6 is 29.4

0 = -4.9t(t-6)

set each equation to 0

-4.9t = 0 or t-6=0

t = 0 or t= 6

it's 6 seconds

after 6 seconds in the air

Step-by-step explanation:

setting 'h' equal to zero will yield when the ball returns to the ground

you can factor out -4.9t to get:

-4.9t(t - 6) = 0

t = 0 (prior to ball being thrown)

t = 6 (this means, after 6 seconds, the height of the ball is back to zero)

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1