Question

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?

Answer 1

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.

Answer 2

i still don't understand????

Answer 3

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

Answer 4

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.

Answer 5

What is the kinetic force