A golfer hits a ball. The ball leaves the club with an initial upward velocity of 40 meters per second. If the ball is 0.25 meters off the ground when it leaves the golf club, about how many seconds will it take for the ball to hit the ground?

Question

anonymous · Answer

Let's get some formula down

x=X0+vt-1/2 g t^2

amistre64 · Answer

h(t) = -meteric gravity units(t^2) +40t +.25

amistre64 · Answer

4.6 or 4.9; cant recall the metric

anonymous · Answer

X0- initial height
v- initial velocity
g- gravity  (9.8)

amistre64 · Answer

h(t) = -4.9t^2 +40t +.25 i believe is good

anonymous · Answer

Set that equal to 0 and solve for t

amistre64 · Answer

I get abt 8.1695..... seconds

amistre64 · Answer

we can plug those into the quadratic formula to get an exact answer;

$$\frac{-40 +\sqrt{40^2-4(-4.9)(.25)}}{2(-4.9)}$$