conservation of energy

[logistic_guy]

A projectile is fired at an upward angle of \(\displaystyle 45^{\circ}\) from the top of a \(\displaystyle 165\)-\(\displaystyle \text{m}\) cliff with a speed of \(\displaystyle 180 \ \text{m/s}\). What will be its speed when it strikes the ground below? (Use conservation of energy)

 

[logistic_guy]

Conservation of energy formula says:

\(\displaystyle E_i = E_f\)

or

\(\displaystyle K_i + U_i = K_f + U_f\)

A mix of kinetic and potential energies. Then we have:

\(\displaystyle \frac{1}{2}mv^2_i + mgh_i = \frac{1}{2}mv^2_f + mgh_f\)

Just plug in the values.

\(\displaystyle \frac{1}{2}(180)^2 + (9.8)(165) = \frac{1}{2}v^2_f + (9.8)(0)\)

This gives:

\(\displaystyle v_f = \textcolor{blue}{189 \ \text{m/s}}\)