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.

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*490 = 245,000 J.

Ek=1/2mv2

I dont really understand the solving.

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

Is u the initial velocity

Good one.👋🏼👋🏼

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

To solve this problem, we can use the principles of conservation of energy.

i) To find the kinetic energy just before impact, we can calculate the difference in potential energy between the drop height and rebound height. The formula for potential energy is given by:

Potential Energy (PE) = mass * gravity * height

Given:
Mass of the ball (m) = 1000g = 1kg (converted from grams to kilograms)
Height of the drop (h1) = 5m
Height of the rebound (h2) = 25m
Acceleration due to gravity (g) = 9.8 m/s²

Step 1: Calculate potential energy at drop height
PE1 = m * g * h1

PE1 = 1kg * 9.8 m/s² * 5m
PE1 = 49 J

Step 2: Calculate potential energy at rebound height
PE2 = m * g * h2

PE2 = 1kg * 9.8 m/s² * 25m
PE2 = 245 J

Step 3: Calculate the change in potential energy
∆PE = PE2 - PE1

∆PE = 245 J - 49 J
∆PE = 196 J

The change in potential energy (∆PE) is equal to the kinetic energy just before impact.

Therefore, the kinetic energy just before impact is 196 Joules.

ii) To find the initial rebound velocity, we can use the principle of conservation of energy. The potential energy at the maximum height is equivalent to the kinetic energy just before impact.

Step 1: Calculate the velocity at maximum height
KE2 = PE2

0.5 * m * v² = ∆PE

0.5 * 1kg * v² = 196 J

Step 2: Solve for v
v² = (2 * 196 J) / 1kg
v² = 392 J/kg

v = √(392 J/kg)
v ≈ 19.8 m/s (rounded to one decimal place)

So, the initial rebound velocity is approximately 19.8 m/s.