a lot of ppl in my science class have trouble with this problem... pls show your work if you know how to solve it.

A juggler throws a ball from height of 0.950 m with a vertical velocity of +4.25 m/s and misses it on the way down. What is its velocity when it hits the ground?

Equate energies at both elevations, assuming no loss due to air resistance.

At h=0.95
Total energies, E1
=PE+KE
=mg(0.95)+(1/2)m(4.25²)

At h=0 (ground)
Total energies, E0
=PE+KE
=mg(0)+(1/2)m(v²)

Equate E0 and E1 and solve for v.

This is similar to the previous problem posted, and g is positive because it acts as a multiplier to convert mass to weight when working with energies.

i still don't understand????

The final energy the ball has is equal to the potential energy change added to the intial kinetic enery.

When an object is thrown upward(vertical), its' final velocity equals

zero:

Vf^2 = Vo^2 + 2gd = 0^2,
(4.25^2) + 2(-9.8)d = 0,
18.06 - 19.6d = 0,
-19.6d = - 18.06,
d = -18.06 / -19.6 = 0.92m.

h = 0.95+0.92 = 1.87m from the ground.

Downward,free-fall:

Vf^2 = Vo^2 + 2*9.8*1.87,
Vf^2 = 0 36.68 = 36.68,
Vf = sqrt(36.68) = 6.06m/s.

What is the kinetic force

To solve this problem, we can use the principles of kinematics. The first step is to identify the given information:

Initial height (h) = 0.950 m
Vertical velocity (v) = +4.25 m/s (upwards)

Now, we need to find the velocity with which the ball hits the ground. To do this, we can utilize the equations of motion. The equation we will use is:

v^2 = u^2 + 2as

where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement

In this case, the ball is thrown upwards, so the acceleration due to gravity acts in the opposite direction, causing the ball to eventually fall downwards. The acceleration due to gravity is approximately -9.8 m/s^2 (negative because it acts downwards).

Now, let's substitute the known values into the equation and solve for the final velocity (v):

v^2 = (4.25 m/s)^2 + 2(-9.8 m/s^2)(0.950 m)
v^2 = 18.0625 m^2/s^2 + (-18.2 m^2/s^2)(0.950 m)
v^2 = 18.0625 m^2/s^2 - 17.29 m^2/s^2
v^2 = 0.7725 m^2/s^2

To find v, we need to take the square root of both sides:

v = sqrt(0.7725 m^2/s^2)
v ≈ 0.88 m/s

Therefore, when the ball hits the ground, its velocity will be approximately 0.88 m/s (downwards).