please show ALL WORK?

Question

bill throws a 10.0g ball straight up from a height of 2.0m The ball strikes the floor with a speed of 7.5m/s. what is the initial speed of the ball?

Answer 1

v = final velocity
u = initial velocity
a = acceleration (gravity)
s = displacement

Falling from maximum height
v^2 = u^2 - 2as
(7.5)^2 = 0 - (2 * -9.81 * s)
56.25 = 19.62s
s = 2.87 metres

Since the ball was thrown up from a height of 2 metres it rises 0.87m

v^2 = u^2 - 2as
0 = u^2 - (2 * 9.81 * 0.87)
0 = u^2 - 17.01
u^2 = 17.01
u = 4.12

Ball was originally thrown up at a velocity of 4.12m/s

note: the mass makes no difference to the velocity

Answer 2

The initial speed of the ball is obviously 0. Bill is holding it, before he throws it. Thus, it is not moving.

Although, if you want to be technical, the Earth is rotating on its axis while the Earth is also traveling around the Sun. The Sun itself is in orbit around the Galactic Center. So, the speed of 0m/s is only relative to Bill and the Earth's surface. It would be greatly different if we change our frame of reference.

Answer 3

distance ball fell is found using

v² = u² + 2gs where g = acceleration due to gravity = 9.8 m/s²
7.5² = 0² + 2*9.8*s (0 as velocity is 0 at top of path)

So s = 56.25/19.6

And so initial distance travelled up = 56.25/19.6 - 2 amd ball thrown with initial velocity u and final veloecity v = 0 at maximum height

So v² = u² - 2gs
0² = u² - 2 *9.8*(56.25/19.6 - 2)

ie u² = 56.25 - 4*9.8 =17.05

u = √17.05
≈ 4.13m/s

Answer 4

v(f)^2=v(i)^2+2ad

v(f)= velocity final (in our case 7.5m/s)
v(i)= velocity initial (in our case the unknown)
a=acceleration(in our case it is 9.8 due to earths gravition)
d=displacement(in our case 2m because the ball travels straight up and then down 2 meters)

plug into the equation

7.5^2=v(i)^2 + 2(9.8)(2)
v(i)^2=17.05
v(i)=4.13m/s

Answer 5

Come on, kid. How you gonna learn anything if you come here to cheat?

Answer 6

10.0+2.0=12
12/7.5=1.6m/s
I dont know is this is right