The greatest height the ball will attain is 3.27 m
<h3>Data obtained from the question</h3>
- Initial velocity (u) = 8 m/s
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the ball can attain can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 8² – (2 × 9.8 × h)
0 = 64 – 19.6h
Collect like terms
0 – 64 = –19.6h
–64 = –19.6h
Divide both side by –19.6
h = –64 / –19.6h
h = 3.27 m
Thus, the greatest height the ball can attain is 3.27 m
Learn more about motion under gravity:
brainly.com/question/13914606
Answer:
u=36.8m/s
Explanation:
because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations
u^2=v^2-2ā*s. where:
u^2 stands for intial velocity
v^2 stands for final velocity
since the cougar skidded to a complete stop the final velocity is zero.
u^2=v^2-2ā*s
u^2=(0)^2 -2(-2.87 m/s^2)*236 m
u^2=0+5.74m/s^2* 236m
u^2=1354.64m^2/s^2
u=√1354.64m^2/s^2
u=36.8m/s (approximate value)
when ever the acceleration is constant you can use one of the following equation to find the required value.
1. v = u + at. (no s)
2. s= 1/2(u+v)t. (no ā)
3. s=ut + 1/2at^2. ( no v)
4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)
Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.
At the top of the first hill, the car's potential energy is
PE = (mass) x (gravity) x (height) .
At the bottom, the car's kinetic energy is
KE = (1/2) (mass) (speed²) .
You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:
KE = 0.9 PE
(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)
Divide each side by (mass):
(0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)
(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)
Divide each side by (0.9):
(0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)
Divide each side by (9.8 m/s²):
Height = (5/9)(4900 m²/s²) / (9.8 m/s²)
= (5 x 4900 m²/s²) / (9 x 9.8 m/s²)
= (24,500 / 88.2) (m²/s²) / (m/s²)
= 277-7/9 meters
(about 911 feet)
Answer:
Image distance is -52.5 cm
Image is virtual and forms on the same side of the lens and upright image is formed.
Explanation:
u = Object distance
v = Image distance
f = Focal length = 35
m = Magnification = 2.5

Lens equation


Image distance is -52.5 cm
Image is virtual and forms on the same side of the lens and upright image is formed.