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.
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?
5 answers
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.
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