Question

A golfer hits a ball from ground level with an initial vertical velocity of 64 ft/s. After how many seconds will the ball hit the ground?

Answer 1

starting at ground level
initial position = y0 = 0, final position = y = 0, gravity = a = -9.80, gravity is negative because it point down in the x-y coordinate system

Answer 2

y= y0 + v0*t + 1/2 a*t^2

v0 = initial velocity

Answer 3

y-y0 =v0*t +1/2 *a *t^2

rearrange so it is easier to see it is a quadratic equation
1/2* a*t^2 +vo*t + (y-y0)

1/2*a =- 4.9
v0= 64
y-y0 = 0

-4.9*t^2 +64*t +0 = 0

plug into quadratic equation and you get
t = 13.06122448
t = 0

t = 0 don't make since so we use 13.06

Answer 4

sorry had to make a correction *
-4.9

Answer 5

another correction *
64 ft /s have to be covert to m/s to have same units a gravity

Answer 6

apply 2nd eq of motion here only