We could approach this problem by using one of the kinematics equations but we should use conservation of energy to solve. Conservation of energy states that the initial total energy of a system is equal to its final total energy. In our case, the system is a rock with an initial speed of $\displaystyle 13\frac{m}{s}$. Choosing the instant at which the rock was thrown at zero height, the rock has only has kinetic energy initially.

$\displaystyle E_i=\frac{1}{2}mv^2$

At maximum height, the velocity is zero (no kinetic energy) and the ball will have only potential energy.

$\displaystyle E_f=mgh$

By conservation of energy, we know that the initial and final energies are equal.

$\displaystyle E_i=E_f$

$\displaystyle \frac{1}{2}mv^2=mgh$

Using this, we can solve for the height.

$\displaystyle h=\frac{v^2}{2g}$

$\displaystyle h=\frac{(13\frac{m}{s})^2}{2(9.8\frac{m}{s^2})}$

$\displaystyle h=8.6\ \text{m}$

This question can be solved using either conservation of energy or linear motion equations.

The ball reaches maximum height when all of the initial kinetic energy is converted to potential energy. From this, we use the relationship:

$\displaystyle KE_i=PE_f$

$\displaystyle \frac{1}{2}mv_i^2=mgh_f$

Mass cancels out from the equation, and we can isolate the velocity variable.

$\displaystyle \frac{1}{2}v_i^2=gh_f$

$\displaystyle v_i=\sqrt{2gh_f}$

We are given the final height and the acceleration from gravity, allowing us to find the initial velocity.

$\displaystyle v_i=\sqrt{2*9.8\frac{m}{s^2}*15m}$

$\displaystyle v_i=17.1 \frac{m}{s}$

We can use conservation of energy to solve this problem. The rock initially has a potential energy, but no kinetic energy. When the rock reaches the ground, it has no potential energy and all kinetic energy. By conservation of energy:

$\displaystyle PE_i=KE_f$

Substitute the equations for potential and kinetic energy.

$\displaystyle mgh_i=\frac{1}{2}mv_f^2$

We can cancel the mass from each term, and isolate the final velocity variable.

$\displaystyle gh_i=\frac{1}{2}v_f^2$

$\displaystyle v_f=\sqrt{2gh_i}$

Finally, use the given value for the height and the acceleration of gravity to solve for the final velocity.

$\displaystyle v=\sqrt{2*9.8\frac{m}{s^2}*100m}$

$\displaystyle v=\sqrt{1960\frac{m^2}{s^2}}=44.3\frac{m}s$

Take the upper end of the spring as “zero.” Since both the spring force and gravitation are conservative forces, the total energy of the block will be conserved as it falls. Knowing the velocity at the upper end of the spring allows calculation of the block’s kinetic energy at that point.

$\displaystyle KE=\frac{1}{2}mv^{2}= (0.5)(5 kg)(20\frac{m}{s})^{2} = 1000 J$

The kinetic energy equals the potential energy it has at its maximum height; since it is falling from rest, it has no kinetic energy at the start. The kinetic energy also equals the maximum spring potential energy the block has at maximum spring compression (where the block has once again come momentarily to rest, and thus has zero kinetic energy).

$\displaystyle PE=mgh=1000J$

$\displaystyle 1000 J = (5 kg)(10\frac{m}{s^{2}})(h)$

$\displaystyle h = 20 m$

Now we will need to look at the spring potential energy.

$\displaystyle E_s=\frac{1}{2}kx^2=1000J$

$\displaystyle 1000J = (0.5)(20\frac{N}{m})(x_{max}^{2})$

$\displaystyle x_{max} = 10 m$

So the total vertical distance traveled is $\displaystyle 20 m$ of fall with an additional $\displaystyle 10 m$ of spring compression for a final distance of $\displaystyle 30 m$.

The bowl described is, fundamentally, a perpetual motion device. When a ball is rolled down the side of a bowl it will gather kinetic energy. As it rolls up the opposite side, this kinetic energy is converted to potential energy until the ball reaches a maximum height. At this point, potential is converted back to kinetic energy. This process would repeat infinitely as long as no energy is lost to the outside environment.

In reality, this machine cannot exist. While a frictionless surface would prevent loss of energy as heat and the vacuum would prevent air resistance, these conditions cannot be met in the real world.

In the given scenario, however, total energy never decreases. The maximum potential energy will always be the same, meaning that the ball will always reach the same maximum height, and the maximum kinetic energy will always be the same, meaning that the ball will always reach the same maximum velocity. Since no energy is lost, both kinetic and potential energy will constantly change in relation to one another, but total mechanical energy will remain forever constant.

This question can be solved by using conservation of energy. Initially, the ball has a displacement of zero, but a non-zero initial velocity. We can say that the initial energy is equal to the total kinetic energy.

$\displaystyle E_1=KE=\frac{1}{2}mv_1^2$

At the peak height of the trajectory, the velocity will be zero and the ball will have a maximum displacement. We can say that the energy at this point is equal to the total potential energy.

$\displaystyle E_2=PE=mgh$

Finally, when the ball lands, its displacement will again be zero and it will have accelerated (due to gravity) to regain the same magnitude of velocity as the initial throw.

$\displaystyle E_3=KE=\frac{1}{2}m(-v_1)^2$

Due to conservation of energy, these terms are all equal. As potential energy increases, kinetic energy decreases, and vice versa. To solve this question, we need to know when the potential energy is equal to half of the total energy.

$\displaystyle E_1=E_2=E_3=E_{max}$

$\displaystyle E_2=mgh\rightarrow \frac{1}{2}E_2=\frac{1}{2}E_{max}=mg(\frac{1}{2}h)$

At exactly half the height, the potential energy will comprise exactly half the total energy. By necessity, the other half will be kinetic energy and the two values will be equal.

The time to travel half the height will be equal to the full flight time divided by four, since acceleration remains constant. The ball will be at this height twice during its trajectory: when it is traveling upward and when it is traveling down. The second time point will be equal to three times this interval.

$\displaystyle \frac{6s}{4}=1.5s$

$\displaystyle 3(1.5s)=4.5s$

At these two times, the kinetic energy and potential energy will have equal values.

Each time the ball bounces, 10% of the kinetic energy is lost. This means that after colliding with the ground, the ball has 90% of its previous potential energy at the top of its flight. To find the total energy after four bounces, we multiply the initial potential energy by $\displaystyle \small 0.9^4$.

The initial potential energy is:

$\displaystyle PE=mgh$

$\displaystyle PE=(0.25kg)(9.8\frac{m}{s^2})(10m)$

$\displaystyle PE=24.5J$

Use our exponential expression to find the remaining energy after four bounces.

$\displaystyle KE_4=24.5*(0.9)^4$

$\displaystyle (24.5J)*(0.66)=16.1J$

Based on conservation of energy, we can find the height that the ball travels with this remaining kinetic energy

$\displaystyle PE_4=16.1J=mgh_4$

$\displaystyle h_4=\frac{PE_4}{mg}$

$\displaystyle h_4=\frac{16.1J}{(0.25kg*9.8\frac{m}{s^2}}=6.6m$