A ball took 0.639s to hit the ground at distance 0.78m from the table. what is the ball's horizontal velocity when it left the table? Please help it’ll be greatly appreciated!!

It leaves the table traveling at 2.35 m/s and falls for 0.639 s.

Explanation:

This is a projectile motion problem. I will use the vertical motion to find the time in flight first, then I will calculate the launch speed. I think it is easiest this way.

Like any object that falls 2.00 m, the time required comes from

Δ

y

=

v

o

Δ

t

+

1

2

a

(

Δ

t

)

2

In this case,

v

o

=

0

,

Δ

y

=

−

2.00

m

(because it is falling) and

a

=

−

9.8

m

s

23

−

2.00

=

−

1

2

(

9.8

)

(

Δ

t

)

2

(

Δ

t

)

2

=

2.00

4.9

=

0.408

Δ

t

=

0.639

s

To travel 1.50 m in 0.639 s, the initial speed must have been

v

o

=

1.50

m

0.639

s

=

2.35

m

s

it takes about 30 days for it to break down

c. newton's third law of motion