Answer :
To determine the minimum initial velocity a ball needs to roll up a 6.5 m high hill, we need to analyze the problem using the principles of physics, specifically the conservation of energy.
The principle of conservation of energy states that in an isolated system, the total energy remains constant. In this problem, we consider the energy transformation from kinetic energy (KE) to gravitational potential energy (PE).
Gravitational Potential Energy at the Top of the Hill:
The potential energy at the top of the hill is given by:
[tex]PE = mgh[/tex]
where:
- [tex]m[/tex] is the mass of the ball,
- [tex]g[/tex] is the acceleration due to gravity ([tex]9.8 \ m/s^2[/tex]),
- [tex]h[/tex] is the height of the hill ([tex]6.5 \ m[/tex]).
Kinetic Energy at the Bottom of the Hill:
The kinetic energy at the bottom of the hill, which needs to be equal to the potential energy at the top for the ball to just reach the top, is given by:
[tex]KE = \frac{1}{2}mv^2[/tex]
where [tex]v[/tex] is the initial velocity of the ball.
Setting Kinetic Energy Equal to Potential Energy:
To find the minimum initial velocity, set the initial kinetic energy equal to the gravitational potential energy:
[tex]\frac{1}{2}mv^2 = mgh[/tex]
The mass [tex]m[/tex] cancels out from both sides:
[tex]\frac{1}{2}v^2 = gh[/tex]
Solve for [tex]v[/tex]:
[tex]v^2 = 2gh[/tex]
[tex]v = \sqrt{2gh}[/tex]
Substitute the Known Values:
[tex]v = \sqrt{2 \times 9.8 \times 6.5}[/tex]
[tex]v = \sqrt{127.4}[/tex]
[tex]v \approx 11.3 \ m/s[/tex]
Therefore, the minimum required initial velocity for the ball to roll all the way up to the top of the hill is approximately [tex]11.3 \ m/s[/tex]. Thus, the correct multiple-choice option is:
C) 11.3 m/s
