The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.
The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)
Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
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Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
