Question
A ball mass 1000g is drop from a height Of 5m and rebounded to a height Of 25 m. Find i it kintentic just before impact ii its intial rebounce velocity and kintentic and show the working.
Answers
Henry
a. V^2 = Vo^2 + 2g*h = 0 + 19.6*5 = 98, V = 9.90 m/s.
Ek = 0.5M*V^2 = 500*98 = 49,000 J.
b. V^2 = Vo^2 + 2g*h.
0 = Vo^2 - 19.6*25, Vo = 22.1 m/s.
Ek =
Ek = 0.5M*V^2 = 500*98 = 49,000 J.
b. V^2 = Vo^2 + 2g*h.
0 = Vo^2 - 19.6*25, Vo = 22.1 m/s.
Ek =
Henry
Ek = 0.5M*V^2 = 500*490 = 245,000 J.
precious
Ek=1/2mv2
Whesu Samuel
I dont really understand the solving.
Ikimot
M=1000g
1000g to kg=1000/1000=1kg
h1=5m
h2=25m
g= 10m/s^2
Kinetic energy before impact
K.E= 1/2mv^2
U=0
V= don't know
So we use the equation of motion
V^2=U^2+2gh
V^2=0^2+2×10×5
V^2=100
V=√100=10m/s
K.E=1/2×1×10×10=50j
(ii) it's initial rebounds velocity and the kinetic energy
Remember that h2=25m
So we use the equation of motion again to get our v
V^2=U^2+2gh2
V^2=0^2+2×10×25
V^2=500
V=√500= 22.14m/s
So the kinetic energy becomes
1/2×1×22.14×22.14=245.0898j
1000g to kg=1000/1000=1kg
h1=5m
h2=25m
g= 10m/s^2
Kinetic energy before impact
K.E= 1/2mv^2
U=0
V= don't know
So we use the equation of motion
V^2=U^2+2gh
V^2=0^2+2×10×5
V^2=100
V=√100=10m/s
K.E=1/2×1×10×10=50j
(ii) it's initial rebounds velocity and the kinetic energy
Remember that h2=25m
So we use the equation of motion again to get our v
V^2=U^2+2gh2
V^2=0^2+2×10×25
V^2=500
V=√500= 22.14m/s
So the kinetic energy becomes
1/2×1×22.14×22.14=245.0898j
Baraqat
Is u the initial velocity
Ossai Princewill
Good one.👋🏼👋🏼
Joy abiodun
I like the solving but i don't understand the initial rebounce velocity and kinetic energy how did they have that 22.14 in the solving