A golf ball rolls off a horizontal cliff with an initial speed of 11.4 m/s. The ball falls a vertical distance of 17.7 m into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed v of the ball just before it strikes the water?

time is 1.90 second

speed is 21.83 m/s

Explanation:

given data

initial speed u = 11.4 m/s

distance s = 17.7 m

to find out

time and speed

solution

we know formula for time that is

time = √2S/g ... 1

here s is distance so put all value

time = √2 (17.7) / 9.8

time = 1.90 second

and

for speed formula is

speed = √ (u1²+u2²) ... 2

here horizontal velocity u1 = 11.4 m/s

and vertical velocity u2 = gt = 9.8*1.90 = 18.62 m/s

so from equation 2

speed = √ (u1²+u2²)

speed = √ (11.4²+18.62²)

speed = 21.83 m/s