Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately
Because energy is wasted either in the form of heat energy or something else (like friction)
It transforms into heat energy
Answer:
(ω₁ / ω₂) = 1.9079
Explanation:
Given
R₁ = 3.59 cm
R₂ = 7.22 cm
m₁ = m₂ = m
K₁ = K₂
We know that
K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²
if
v₁ = ω₁*R₁
and
I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²
∴ K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² <em>(I)</em>
then
K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²
if
v₂ = ω₂*R₂
and
I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²
∴ K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂² <em>(II)</em>
<em>∵ </em>K₁ = K₂
⇒ 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²
⇒ ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²
⇒ (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²
⇒ (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)
⇒ (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²
⇒ (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)
⇒ (ω₁ / ω₂) = 1.9079
