Agolf 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

there is no correct list among the choices you've provided.

the list might look something like this:

-- baby human

-- piccolo

-- flute

-- soprano singer

-- trumpet

-- tenor singer

-- trombone

-- baritone singer

-- baritone horn

-- bass singer

-- tuba

since i had to make up the list myself, i made it easy on myself, and i didn't try to shuffle the string instruments and woodwinds into it too.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s