Answer:
a)T total = 2*Voy/(g*sin( α ))
b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)
α = 90º, T total=2*Voy/g
Explanation:
Voy=Vo*sinα
- Time to reach the maximal height :
Kinematics equation: Vfy=Voy-at
a=g*sinα ; g is gravity
if Vfy=0 ⇒ t=T ; time to reach the maximal height
so:
0=Voy-g*sin( α )*T
T=Voy/(g*sin( α ))
- Time required to return to the starting point:
After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.
So T total= 2T = 2*Voy/(g*sin( α ))
The particle goes totally horizontal, goes away indefinitely
T total= 2*Voy/(g*sin( α )) ≅∞
T total=2*Voy/g
Answer: a stone thrown into a lake
Explanation:
The option that exhibits parabolic motion is a stone that is thrown into a lake.
Projectile motion is a form of motion that is seen when a particular object is thrown towards the surface fo the Earth. Such objects are acted upon by gravity.
Answer:
Because the electricity flows through and creates static bonds around the metal case which creates a bond with other fields that protects it.
Explanation:
Answer:
The skier’s speed on the two level stretches are 22.13 m/s and 27.29 m/s.
Explanation:
Given that,
Slope down = 32°
Height = 25 m
Before leveling,
Slope down = 20°
Height = 38 m
We need to calculate the skier’s speed on the two level stretches
Using formula of energy
For first stretch,
Height = 25 m
Put the value into the formula
For second stretch,
Height = 38 m
Put the value into the formula
Hence, The skier’s speed on the two level stretches are 22.13 m/s and 27.29 m/s.
Answer:
2.8 seconds
Explanation:
Given:
Δy = 38 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(38 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 2.8 s
