Kiera climbs up a ladder to sit at the top of a slide $2 \mathrm{m}$ above the ground. Her potential energy increases by $1280 \mathrm{J}$.

Kiera then slides down the slide, starting from rest. The slide is modelled as a slope at an angle $\theta$ to the horizontal. The resistance force is a constant $20 \mathrm{N}$. The work done against resistance by Kiera when she is sliding is $80 \mathrm{J}$. Find Kiera's speed when she reaches the bottom of the slide. (Use: $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

A ramp is inclined at an angle ${\sin }^{-1}0.1$ to the horizontal. A box of mass $40 \mathrm{kg}$ is projected up the ramp with initial speed $5 \mathrm{m}{ \mathrm{s}}^{-1}$. The coefficient of friction between the ramp and the box is $0.05$, and no other resistance forces act.

Find the acceleration of the box, stating its direction. The box comes to rest when it reaches the top of the ramp. (Use: $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

A ramp is inclined at an angle ${\sin }^{-1}0.1$ to the horizontal. A box of mass $40 \mathrm{kg}$ is projected up the ramp with initial speed $5 \mathrm{m}{ \mathrm{s}}^{-1}$. The coefficient of friction between the ramp and the box is $0.05$, and no other resistance forces act.

Find the length of the ramp.

A ramp is inclined at an angle ${\sin }^{-1}0.1$ to the horizontal. A box of mass $40 \mathrm{kg}$ is projected up the ramp with initial speed $5 \mathrm{m}{ \mathrm{s}}^{-1}$. The coefficient of friction between the ramp and the box is $0.05$, and no other resistance forces act.

Find the gain in the potential energy of the box.

The total mechanical energy is the sum of the kinetic energy and the potential energy. 

A ramp is inclined at an angle ${\sin }^{-1}0.1$ to the horizontal. A box of mass $40 \mathrm{kg}$ is projected up the ramp with initial speed $5 \mathrm{m}{ \mathrm{s}}^{-1}$. The coefficient of friction between the ramp and the box is $0.05$, and no other resistance forces act.

Show that the overall loss in the mechanical energy of the box is $166 \mathrm{J}$.

Jack has mass $70 \mathrm{kg}$. He works as a 'human cannon ball'. Jack is projected with speed $12 \mathrm{m}{ \mathrm{s}}^{-1}$ at an angle of $45°$ above the horizontal. He lands on a trampoline when the angle between his flight and the horizontal is $50°$. Model Jack as a particle with no air resistance.

Explain why the horizontal component of Jack's velocity is constant.

Jack has mass $70 \mathrm{kg}$. He works as a 'human cannon ball'. Jack is projected with speed $12 \mathrm{m}{ \mathrm{s}}^{-1}$ at an angle of $45°$ above the horizontal. He lands on a trampoline when the angle between his flight and the horizontal is $50°$. Model Jack as a particle with no air resistance.

Find Jack's speed when he hits the trampoline.

Jack has mass $70 \mathrm{kg}$. He works as a 'human cannon ball'. Jack is projected with speed $12 \mathrm{m}{ \mathrm{s}}^{-1}$ at an angle of $45°$ above the horizontal. He lands on a trampoline when the angle between his flight and the horizontal is $50°$. Model Jack as a particle with no air resistance.

Find the kinetic energy gained during the flight.

The gain in Jack's kinetic energy equals the loss in his gravitational potential energy.

Jack has mass $70 \mathrm{kg}$. He works as a 'human cannon ball'. Jack is projected with speed $12 \mathrm{m}{ \mathrm{s}}^{-1}$ at an angle of $45°$ above the horizontal. He lands on a trampoline when the angle between his flight and the horizontal is $50°$. Model Jack as a particle with no air resistance.

Find the difference in height between the mouth of the cannon and the trampoline. (Use: $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)