Technically, this delivers a lot of energy into the Earth, but it’s
spread out over a large enough area that it doesn’t do much more than
leave footprints in a lot of gardens. A slight pulse of pressure spreads
through the North American continental crust and dissipates with little
effect. The sound of all those feet hitting the ground creates a loud,
drawn-out roar which lasts many seconds.
The speed of the toy when it hits the ground is 2.97 m/s.
The given parameters;
- mass of the toy, m = 0.1 kg
- the maximum height reached by the, h = 0.45 m
The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.
P.E = K.E

Substitute the given values and solve the speed;

Thus, the speed of the toy when it hits the ground is 2.97 m/s.
Learn more here: brainly.com/question/7562874
Answer:
The increase in potential energy of the ball is 115.82 J
Explanation:
Conceptual analysis
Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:
U = m × g × h
U: Potential Energy in Joules (J)
m: mass in kg
g: acceleration due to gravity in m/s²
h: height in m
Equivalences
1 kg = 1000 g
1 ft = 0.3048 m
1 N = 1 (kg×m)/s²
1 J = N × m
Known data




Problem development
ΔU: Potential energy change
ΔU = U₂ - U₁
U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁
U₂ - U₁ = mₓg(h₂ - h₁)

The increase in potential energy of the ball is 115.82 J
P1v1/t1 = p2v2/t2
p1=p2, v1=.2, t1=333, t2=533
we can find v2 from this
be aware, temperature must be in Kelvin.
Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles